Question

CaCO3 + 2HCl > CaCl2 + H2O + CO2

a.) 45.0 g CaCO3 are added to 1.25 L of a 25.7% (w/v) solution of HCl. Calculate the theoretical yield of CO2 formed.

b.) Calculate the mass of CaCO3 and volume of HCl solution required to produce 25.0 g of CO2. The reaction is 78.0% efficient.

Answer 1

Answer on Question #46059, Chemistry, Other

Task:

$\mathrm{CaCO_3 + 2HCl > CaCl_2 + H_2O + CO_2}$

a.) 45.0 g CaCO₃ are added to 1.25 L of a 25.7% (w/v) solution of HCl. Calculate the theoretical yield of CO₂ formed.

b.) Calculate the mass of CaCO₃ and volume of HCl solution required to produce 25.0 g of CO₂. The reaction is 78.0% efficient.

Answer:

a) $v = \frac{m}{M}$

$$v \left(\mathrm{CaCO_3}\right) = \frac{45}{100} = 0.45 \mathrm{~mol}$$ $$\mathrm{M} \left(\mathrm{CaCO_3}\right) = 100 \mathrm{~g/mol}$$ $$\text{Volume percent} = \frac{\text{Weight of solute (g)}}{\text{Volume of solution (mL)}} \cdot 100\%$$ $$m(HCl) = \frac{1250 \cdot 25.7}{100} = 321.3 \mathrm{~g}$$ $$\mathrm{M(HCl)} = 36.5 \mathrm{~g/mol}$$ $$v(HCl) = \frac{321.5}{36.5} = 8.8 \mathrm{~mol}$$

From these calculations we can say, that HCl is in excess and CaCO₃ is the limiting reagent.

$$v \left(\mathrm{CO_2}\right) = v \left(\mathrm{CaCO_3}\right) = 0.45 \mathrm{~mol}$$ $$v \left(\mathrm{CO_2}\right) = \frac{V \left(\mathrm{CO_2}\right)}{22.4}$$ $$V \left(\mathrm{CO_2}\right) = v \left(\mathrm{CO_2}\right) \cdot 22.4 = 0.45 \cdot 22.4 = 10.08 \mathrm{l}$$

b) 25.0 g of CO₂ is only 78% of the reaction product. So, the 100% amount will be:

$$m(\mathrm{CO_2}) = \frac{25 \cdot 100}{78} = 32.05 \mathrm{~g}$$ $$v(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO_2})} = \frac{32.05}{44} = 0.73 \mathrm{~mol}$$

According to the equation:

$$v(HCl) = 2 \cdot v(\mathrm{CO_2}) = 2 \cdot 0.73 \mathrm{~mol} = 1.46 \mathrm{~mol}$$ $$v(\mathrm{NaCO_3}) = v(\mathrm{CO_2}) = 0.73 \mathrm{~mol}$$ $$v = \frac{m}{M}$$ $$m(HCl) = v(HCl) \cdot M(HCl) = 1.46 \cdot 36.5 = 53.3 \mathrm{~g}$$

We will use the concentration 25.7% (w/v). So, the amount of this solution will be:

$$V(HCl) = \frac{53.3 \cdot 100}{25.7} = 207 \mathrm{~mL}$$ $$m(\mathrm{CaCO_3}) = v(\mathrm{CaCO_3}) \cdot M(\mathrm{CaCO_3}) = 0.73 \cdot 100 = 73 \mathrm{~g}$$

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