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What is the percent yield of a reaction in which 45.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 9.60 mL of water (d = 1.00 g/mL)?

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.92: Overall Percent Yield

For successive reactions, the overall percent yield is the product of the individual percent yields (expressed as decimals).

  1. Convert individual percent yields to decimals.

    Yield of D ⟶\longrightarrow E = 48%=0.4848\% = 0.48
    Yield of E ⟶\longrightarrow F = 73%=0.7373\% = 0.73

  2. Calculate the overall yield.

    Overall Yield=Yield (D→E)×Yield (E→F)\text{Overall Yield} = \text{Yield (D} \to \text{E)} \times \text{Yield (E} \to \text{F)} Overall Yield=0.48×0.73=0.3504\text{Overall Yield} = 0.48 \times 0.73 = 0.3504

  3. Convert the overall yield back to a percentage.

    Overall Percent Yield=0.3504×100%=35.04%\text{Overall Percent Yield} = 0.3504 \times 100\% = 35.04\%

    Rounding to two significant figures:

    \boxed{\text{35%}}

    .93: Percent Yield of Tungsten(VI) Oxide Reaction

    The balanced chemical equation for the reaction is:

    WO3(s)+3H2(g)⟶W(s)+3H2O(l)\text{WO}_3(s) + 3\text{H}_2(g) \longrightarrow \text{W}(s) + 3\text{H}_2\text{O}(l)

    Given:

    • Mass of WO3=45.5 g\text{WO}_3 = 45.5 \text{ g}
    • Actual yield of H2O=9.60 mL\text{H}_2\text{O} = 9.60 \text{ mL}
    • Density of H2O=1.00 g/mL\text{H}_2\text{O} = 1.00 \text{ g/mL}

  4. Calculate the molar masses of WO3\text{WO}_3 and H2O\text{H}_2\text{O}.

    • Molar mass of WO3=183.84 g/mol (W)+(3×16.00 g/mol (O))=231.84 g/mol\text{WO}_3 = 183.84 \text{ g/mol (W)} + (3 \times 16.00 \text{ g/mol (O)}) = 231.84 \text{ g/mol}
    • Molar mass of H2O=(2×1.008 g/mol (H))+16.00 g/mol (O)=18.016 g/mol\text{H}_2\text{O} = (2 \times 1.008 \text{ g/mol (H)}) + 16.00 \text{ g/mol (O)} = 18.016 \text{ g/mol}

  5. Convert the actual yield of water from volume to mass.

    Actual mass of H2O=volume×density=9.60 mL×1.00 g/mL=9.60 g\text{Actual mass of H}_2\text{O} = \text{volume} \times \text{density} = 9.60 \text{ mL} \times 1.00 \text{ g/mL} = 9.60 \text{ g}

  6. Calculate the moles of WO3\text{WO}_3.

    moles of WO3=45.5 g231.84 g/mol=0.19625 mol\text{moles of WO}_3 = \frac{45.5 \text{ g}}{231.84 \text{ g/mol}} = 0.19625 \text{ mol}

  7. Calculate the theoretical moles of H2O\text{H}_2\text{O} produced.

    From the balanced equation, 1 mol of WO31 \text{ mol of WO}_3 produces 3 mol of H2O3 \text{ mol of H}_2\text{O}.

    theoretical moles of H2O=0.19625 mol WO3×3 mol H2O1 mol WO3=0.58875 mol H2O\text{theoretical moles of H}_2\text{O} = 0.19625 \text{ mol WO}_3 \times \frac{3 \text{ mol H}_2\text{O}}{1 \text{ mol WO}_3} = 0.58875 \text{ mol H}_2\text{O}

  8. Calculate the theoretical mass (yield) of H2O\text{H}_2\text{O}.

    theoretical mass of H2O=0.58875 mol×18.016 g/mol=10.607 g\text{theoretical mass of H}_2\text{O} = 0.58875 \text{ mol} \times 18.016 \text{ g/mol} = 10.607 \text{ g}

  9. Calculate the percent yield.

    Percent Yield=Actual YieldTheoretical Yield×100%\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% Percent Yield=9.60 g10.607 g×100%=90.506%\text{Percent Yield} = \frac{9.60 \text{ g}}{10.607 \text{ g}} \times 100\% = 90.506\%

    Rounding to three significant figures:

    \boxed{\text{90.5%}}

    .94: Percent Yield of Phosphorus Trichloride Reaction

    The balanced chemical equation for the reaction is:

    PCl3(l)+3H2O(l)⟶3HCl(aq)+H3PO3(aq)\text{PCl}_3(l) + 3\text{H}_2\text{O}(l) \longrightarrow 3\text{HCl}(aq) + \text{H}_3\text{PO}_3(aq)

    Given:

    • Mass of PCl3=200. g\text{PCl}_3 = 200. \text{ g}
    • Actual yield of HCl=128 g\text{HCl} = 128 \text{ g}
    • Water is in excess.

  10. Calculate the molar masses of PCl3\text{PCl}_3 and HCl\text{HCl}.

    • Molar mass of PCl3=30.97 g/mol (P)+(3×35.45 g/mol (Cl))=137.32 g/mol\text{PCl}_3 = 30.97 \text{ g/mol (P)} + (3 \times 35.45 \text{ g/mol (Cl)}) = 137.32 \text{ g/mol}
    • Molar mass of HCl=1.008 g/mol (H)+35.45 g/mol (Cl)=36.458 g/mol\text{HCl} = 1.008 \text{ g/mol (H)} + 35.45 \text{ g/mol (Cl)} = 36.458 \text{ g/mol}

  11. Calculate the moles of PCl3\text{PCl}_3.

    moles of PCl3=200. g137.32 g/mol=1.45645 mol\text{moles of PCl}_3 = \frac{200. \text{ g}}{137.32 \text{ g/mol}} = 1.45645 \text{ mol}

  12. Calculate the theoretical moles of HCl\text{HCl} produced.

    From the balanced equation, 1 mol of PCl31 \text{ mol of PCl}_3 produces 3 mol of HCl3 \text{ mol of HCl}.

    theoretical moles of HCl=1.45645 mol PCl3×3 mol HCl1 mol PCl3=4.36935 mol HCl\text{theoretical moles of HCl} = 1.45645 \text{ mol PCl}_3 \times \frac{3 \text{ mol HCl}}{1 \text{ mol PCl}_3} = 4.36935 \text{ mol HCl}

  13. Calculate the theoretical mass (yield) of HCl\text{HCl}.

    theoretical mass of HCl=4.36935 mol×36.458 g/mol=159.29 g\text{theoretical mass of HCl} = 4.36935 \text{ mol} \times 36.458 \text{ g/mol} = 159.29 \text{ g}

  14. Calculate the percent yield.

    Percent Yield=Actual YieldTheoretical Yield×100%\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% Percent Yield=128 g159.29 g×100%=80.356%\text{Percent Yield} = \frac{128 \text{ g}}{159.29 \text{ g}} \times 100\% = 80.356\%

    Rounding to three significant figures:

    \boxed{\text{80.4%}}

    .95: Mass of Chloromethane Produced

    The balanced chemical equation for the reaction is:

    CH4(g)+Cl2(g)⟶CH3Cl(g)+HCl(g)\text{CH}_4(g) + \text{Cl}_2(g) \longrightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g)

    Given:

    • Mass of CH4=20.5 g\text{CH}_4 = 20.5 \text{ g}
    • Mass of Cl2=45.0 g\text{Cl}_2 = 45.0 \text{ g}
    • Percent yield = 75.0%75.0\%

  15. Calculate the molar masses of CH4\text{CH}_4, Cl2\text{Cl}_2, and CH3Cl\text{CH}_3\text{Cl}.

    • Molar mass of CH4=12.01 g/mol (C)+(4×1.008 g/mol (H))=16.042 g/mol\text{CH}_4 = 12.01 \text{ g/mol (C)} + (4 \times 1.008 \text{ g/mol (H)}) = 16.042 \text{ g/mol}
    • Molar mass of Cl2=2×35.45 g/mol (Cl)=70.90 g/mol\text{Cl}_2 = 2 \times 35.45 \text{ g/mol (Cl)} = 70.90 \text{ g/mol}
    • Molar mass of CH3Cl=12.01 g/mol (C)+(3×1.008 g/mol (H))+35.45 g/mol (Cl)=50.484 g/mol\text{CH}_3\text{Cl} = 12.01 \text{ g/mol (C)} + (3 \times 1.008 \text{ g/mol (H)}) + 35.45 \text{ g/mol (Cl)} = 50.484 \text{ g/mol}

  16. Calculate the moles of each reactant.

    moles of CH4=20.5 g16.042 g/mol=1.2779 mol\text{moles of CH}_4 = \frac{20.5 \text{ g}}{16.042 \text{ g/mol}} = 1.2779 \text{ mol} moles of Cl2=45.0 g70.90 g/mol=0.63469 mol\text{moles of Cl}_2 = \frac{45.0 \text{ g}}{70.90 \text{ g/mol}} = 0.63469 \text{ mol}

  17. Determine the limiting reactant.

    From the balanced equation, 1 mol of CH41 \text{ mol of CH}_4 reacts with 1 mol of Cl21 \text{ mol of Cl}_2.
    Since 0.63469 mol of Cl20.63469 \text{ mol of Cl}_2 is less than 1.2779 mol of CH41.2779 \text{ mol of CH}_4, Cl2\text{Cl}_2 is the limiting reactant.

  18. Calculate the theoretical moles of CH3Cl\text{CH}_3\text{Cl} produced from the limiting reactant.

    From the balanced equation, 1 mol of Cl21 \text{ mol of Cl}_2 produces 1 mol of CH3Cl1 \text{ mol of CH}_3\text{Cl}.

    theoretical moles of CH3Cl=0.63469 mol Cl2×1 mol CH3Cl1 mol Cl2=0.63469 mol CH3Cl\text{theoretical moles of CH}_3\text{Cl} = 0.63469 \text{ mol Cl}_2 \times \frac{1 \text{ mol CH}_3\text{Cl}}{1 \text{ mol Cl}_2} = 0.63469 \text{ mol CH}_3\text{Cl}

  19. Calculate the theoretical mass (yield) of CH3Cl\text{CH}_3\text{Cl}.

    theoretical mass of CH3Cl=0.63469 mol×50.484 g/mol=32.049 g\text{theoretical mass of CH}_3\text{Cl} = 0.63469 \text{ mol} \times 50.484 \text{ g/mol} = 32.049 \text{ g}

  20. Calculate the actual mass of CH3Cl\text{CH}_3\text{Cl} using the given percent yield.

    Actual Yield=Theoretical Yield×Percent Yield100%\text{Actual Yield} = \text{Theoretical Yield} \times \frac{\text{Percent Yield}}{100\%} Actual Yield=32.049 g×75.0%100%=32.049 g×0.750=24.03675 g\text{Actual Yield} = 32.049 \text{ g} \times \frac{75.0\%}{100\%} = 32.049 \text{ g} \times 0.750 = 24.03675 \text{ g}

    Rounding to three significant figures:

    24.0 g\boxed{\text{24.0 g}}

    .96: Mass of Calcium Nitride Produced

    The balanced chemical equation for the reaction is:

    3Ca(s)+N2(g)⟶Ca3N2(s)3\text{Ca}(s) + \text{N}_2(g) \longrightarrow \text{Ca}_3\text{N}_2(s)

    Given:

    • Mass of Ca=56.8 g\text{Ca} = 56.8 \text{ g}
    • Mass of N2=30.5 g\text{N}_2 = 30.5 \text{ g}
    • Percent yield = 93.0%93.0\%

  21. Calculate the molar masses of Ca\text{Ca}, N2\text{N}_2, and Ca3N2\text{Ca}_3\text{N}_2.

    • Molar mass of Ca=40.08 g/mol\text{Ca} = 40.08 \text{ g/mol}
    • Molar mass of N2=2×14.01 g/mol (N)=28.02 g/mol\text{N}_2 = 2 \times 14.01 \text{ g/mol (N)} = 28.02 \text{ g/mol}
    • Molar mass of Ca3N2=(3×40.08 g/mol (Ca))+(2×14.01 g/mol (N))=120.24+28.02=148.26 g/mol\text{Ca}_3\text{N}_2 = (3 \times 40.08 \text{ g/mol (Ca)}) + (2 \times 14.01 \text{ g/mol (N)}) = 120.24 + 28.02 = 148.26 \text{ g/mol}

  22. Calculate the moles of each reactant.

    moles of Ca=56.8 g40.08 g/mol=1.4172 mol\text{moles of Ca} = \frac{56.8 \text{ g}}{40.08 \text{ g/mol}} = 1.4172 \text{ mol} moles of N2=30.5 g28.02 g/mol=1.0885 mol\text{moles of N}_2 = \frac{30.5 \text{ g}}{28.02 \text{ g/mol}} = 1.0885 \text{ mol}

  23. Determine the limiting reactant.

    From the balanced equation, 3 mol of Ca3 \text{ mol of Ca} reacts with 1 mol of N21 \text{ mol of N}_2.
    To react all 1.4172 mol of Ca1.4172 \text{ mol of Ca}, we need:

    1.4172 mol Ca×1 mol N23 mol Ca=0.4724 mol N21.4172 \text{ mol Ca} \times \frac{1 \text{ mol N}_2}{3 \text{ mol Ca}} = 0.4724 \text{ mol N}_2

    Since we have 1.0885 mol of N21.0885 \text{ mol of N}_2 (which is more than 0.4724 mol0.4724 \text{ mol} needed), calcium (Ca\text{Ca}) is the limiting reactant.

  24. Calculate the theoretical moles of Ca3N2\text{Ca}_3\text{N}_2 produced from the limiting reactant.

    From the balanced equation, 3 mol of Ca3 \text{ mol of Ca} produces 1 mol of Ca3N21 \text{ mol of Ca}_3\text{N}_2.

    theoretical moles of Ca3N2=1.4172 mol Ca×1 mol Ca3N23 mol Ca=0.4724 mol Ca3N2\text{theoretical moles of Ca}_3\text{N}_2 = 1.4172 \text{ mol Ca} \times \frac{1 \text{ mol Ca}_3\text{N}_2}{3 \text{ mol Ca}} = 0.4724 \text{ mol Ca}_3\text{N}_2

  25. Calculate the theoretical mass (yield) of Ca3N2\text{Ca}_3\text{N}_2.

    theoretical mass of Ca3N2=0.4724 mol×148.26 g/mol=70.03 g\text{theoretical mass of Ca}_3\text{N}_2 = 0.4724 \text{ mol} \times 148.26 \text{ g/mol} = 70.03 \text{ g}

  26. Calculate the actual mass of Ca3N2\text{Ca}_3\text{N}_2 using the given percent yield.

    Actual Yield=Theoretical Yield×Percent Yield100%\text{Actual Yield} = \text{Theoretical Yield} \times \frac{\text{Percent Yield}}{100\%} Actual Yield=70.03 g×93.0%100%=70.03 g×0.930=65.1279 g\text{Actual Yield} = 70.03 \text{ g} \times \frac{93.0\%}{100\%} = 70.03 \text{ g} \times 0.930 = 65.1279 \text{ g}

    Rounding to three significant figures:

    65.1 g\boxed{\text{65.1 g}}
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