What is the gas produced from calcium carbide and water, its balanced equation, and the mass of slaked lime…

Question

What is the gas produced from calcium carbide and water, its balanced equation, and the mass of slaked lime from 2.5 g of calcium carbide?

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to your questions:

1.a)

i) The gas obtained is acetylene with the formula $\text{C}_2\text{H}_2$ .
ii) The balanced chemical equation is: $\text{CaC}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \longrightarrow \text{Ca(OH)}_2(\text{aq}) + \text{C}_2\text{H}_2(\text{g})$
iii) Calculate the mass of slaked lime obtained from $2.5 \text{ g}$ of calcium carbide. Given molar masses: $\text{Ca}=40 \text{ g/mol}$ , $\text{C}=12 \text{ g/mol}$ , $\text{H}=1 \text{ g/mol}$ , $\text{O}=16 \text{ g/mol}$ .

Molar mass of $\text{CaC}_2 = 40 + 2(12) = 64 \text{ g/mol}$  
Molar mass of $\text{Ca(OH)}_2 = 40 + 2(16+1) = 74 \text{ g/mol}$

$$
\text{Moles of CaC}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.5 \text{ g}}{64 \text{ g/mol}} = 0.0390625 \text{ mol}
$$

Use the mole ratio from the balanced equation ( $\text{CaC}_2 : \text{Ca(OH)}_2 = 1:1$ ) to find moles of $\text{Ca(OH)}_2$ .
$$ \text{Moles of Ca(OH)}_2 = 0.0390625 \text{ mol} $$
Calculate the mass of $\text{Ca(OH)}_2$ produced.
$$ \text{Mass of Ca(OH)}_2 = \text{Moles} \times \text{Molar mass} = 0.0390625 \text{ mol} \times 74 \text{ g/mol} = 2.890625 \text{ g} $$ The mass of slaked lime obtained is $\boxed{\text{2.89 g}}$.
1.b)
- i) Calculate the number of Faradays used. Given: Current $I = 2 \text{ A}$ , Time $t = 15 \text{ minutes}$ .

$$
t = 15 \text{ min} \times 60 \text{ s/min} = 900 \text{ s}
$$

$$
Q = I \times t = 2 \text{ A} \times 900 \text{ s} = 1800 \text{ C}
$$

Calculate the number of Faradays.
(Faraday constant $F \approx 96500 \text{ C/mol}$ ) $\text{Number of Faradays} = \frac{Q}{F} = \frac{1800 \text{ C}}{96500 \text{ C/mol}} = 0.01865 \text{ mol}$ The number of Faradays used is $\boxed{\text{0.0187 mol}}$ .
- ii) Calculate the number of moles of Lithium atoms produced. The reaction at the cathode for lithium hydride electrolysis is: $\text{Li}^+ + \text{e}^- \longrightarrow \text{Li}$ . From the equation, $1 \text{ mole of electrons}$ produces $1 \text{ mole of Li}$ .
Moles of electrons passed = Number of Faradays.
$$ \text{Moles of electrons} = 0.01865 \text{ mol} $$
Calculate moles of Lithium produced.
$$ \text{Moles of Li} = \text{Moles of electrons} = 0.01865 \text{ mol} $$ The number of moles of Lithium atoms produced is $\boxed{\text{0.0187 mol}}$.
- iii) Calculate the volume of gas produced at r.t.p. The gas produced at the anode from lithium hydride electrolysis is hydrogen: $2\text{H}^- \longrightarrow \text{H}_2 + 2\text{e}^-$ . From the equation, $2 \text{ moles of electrons}$ produce $1 \text{ mole of H}_2$ .

$$
\text{Moles of H}_2 = \frac{\text{Moles of electrons}}{2} = \frac{0.01865 \text{ mol}}{2} = 0.009325 \text{ mol}
$$

Calculate the volume of $\text{H}_2$ gas at r.t.p.
(Molar volume at r.t.p = $2400 cm^3/\text{

What is the gas produced from calcium carbide and water, its balanced equation, and the mass of slaked lime from 2.5 g of calcium carbide?