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What is the mass % of nitrogen in urea, in arginine, and in or-nithine? (b) How many grams of nitrogen can be excreted as urea when 135.2 g of ornithine is produced?

Chemistry

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Answer

1.77 g\text{1.77 g}

Step-by-step solution

5 steps

To determine the grams of chromite (FeCr2_2O4_4) needed, we will follow these steps:

  1. Calculate the mass of lead(II) chromate (PbCrO4_4) required for 1 kilogram of paint.
  2. Convert the mass of PbCrO4_4 to moles using its molar mass.
  3. Use the mole ratio between PbCrO4_4 and K2_2CrO4_4 (from the replacement reaction) to find moles of K2_2CrO4_4.
  4. Use the mole ratio from the balanced chemical equation to find moles of FeCr2_2O4_4.
  5. Convert moles of FeCr2_2O4_4 to grams using its molar mass.

First, calculate the molar masses:
Molar mass of PbCrO4_4:
Pb: 207.2 g/mol207.2 \text{ g/mol}
Cr: 52.00 g/mol52.00 \text{ g/mol}
O: 16.00 g/mol16.00 \text{ g/mol}

Molar mass of PbCrO4=207.2+52.00+(4×16.00)=323.2 g/mol\text{Molar mass of PbCrO}_4 = 207.2 + 52.00 + (4 \times 16.00) = 323.2 \text{ g/mol}

Molar mass of FeCr2_2O4_4:
Fe: 55.845 g/mol55.845 \text{ g/mol}
Cr: 52.00 g/mol52.00 \text{ g/mol}
O: 16.00 g/mol16.00 \text{ g/mol}

Molar mass of FeCr2O4=55.845+(2×52.00)+(4×16.00)=223.845 g/mol\text{Molar mass of FeCr}_2\text{O}_4 = 55.845 + (2 \times 52.00) + (4 \times 16.00) = 223.845 \text{ g/mol}
  1. Calculate the mass of PbCrO4_4 in 1 kg of paint.

    A kilogram of paint is 1000 g1000 \text{ g}. The paint is 0.511%0.511\% PbCrO4_4 by mass.

    Mass of PbCrO4=1000 g paint×0.511 g PbCrO4100 g paint=5.11 g PbCrO4\text{Mass of PbCrO}_4 = 1000 \text{ g paint} \times \frac{0.511 \text{ g PbCrO}_4}{100 \text{ g paint}} = 5.11 \text{ g PbCrO}_4
  2. Convert the mass of PbCrO4_4 to moles.

    Moles of PbCrO4=5.11 g PbCrO4×1 mol PbCrO4323.2 g PbCrO4=0.0158106 mol PbCrO4\text{Moles of PbCrO}_4 = 5.11 \text{ g PbCrO}_4 \times \frac{1 \text{ mol PbCrO}_4}{323.2 \text{ g PbCrO}_4} = 0.0158106 \text{ mol PbCrO}_4
  3. Determine moles of K2_2CrO4_4.

    The problem states that "Lead(II) ion then replaces the K+^+ ion" to form PbCrO4_4. This implies a 1:1 mole ratio between K2_2CrO4_4 and PbCrO4_4.

    Moles of K2CrO4=Moles of PbCrO4=0.0158106 mol K2CrO4\text{Moles of K}_2\text{CrO}_4 = \text{Moles of PbCrO}_4 = 0.0158106 \text{ mol K}_2\text{CrO}_4
  4. Determine moles of FeCr2_2O4_4 from the balanced equation.

    The balanced equation is:

    4FeCr2O4(s)+8K2CO3(aq)+7O2(g)⟶2Fe2O3(s)+8K2CrO4(aq)+8CO2(g)4\text{FeCr}_2\text{O}_4(s) + 8\text{K}_2\text{CO}_3(aq) + 7\text{O}_2(g) \longrightarrow 2\text{Fe}_2\text{O}_3(s) + 8\text{K}_2\text{CrO}_4(aq) + 8\text{CO}_2(g)

    From the equation, 4 moles of FeCr2_2O4_4 produce 8 moles of K2_2CrO4_4. The mole ratio is 4:84:8, or 1:21:2.

    Moles of FeCr2O4=0.0158106 mol K2CrO4×4 mol FeCr2O48 mol K2CrO4=0.0079053 mol FeCr2O4\text{Moles of FeCr}_2\text{O}_4 = 0.0158106 \text{ mol K}_2\text{CrO}_4 \times \frac{4 \text{ mol FeCr}_2\text{O}_4}{8 \text{ mol K}_2\text{CrO}_4} = 0.0079053 \text{ mol FeCr}_2\text{O}_4
  5. Convert moles of FeCr2_2O4_4 to grams.

    Mass of FeCr2O4=0.0079053 mol FeCr2O4×223.845 g FeCr2O41 mol FeCr2O4=1.7695 g FeCr2O4\text{Mass of FeCr}_2\text{O}_4 = 0.0079053 \text{ mol FeCr}_2\text{O}_4 \times \frac{223.845 \text{ g FeCr}_2\text{O}_4}{1 \text{ mol FeCr}_2\text{O}_4} = 1.7695 \text{ g FeCr}_2\text{O}_4

    Rounding to three significant figures, the mass of chromite needed is 1.77 g1.77 \text{ g}.

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