Question

In CaCo3 + HCl = CaCl2 + H2O +CO3, how many liters of carbon dioxide gas, measured at STP, can be obtained from 45.0 g of calcium carbonate?
A) 19.8 L


B) 10.1 L


C) 22.4 L


D)2.28 L

Answer 1

Answer:

B) 10.1 L

Explanation:

Hello,

In this case, for the given chemical reaction which should be corrected as shown below:

$CaCO_3 + HCl \rightarrow CaCl_2 + H_2O +CO_2$

Since 45.0g of calcium carbonate are used, the produced moles of carbon dioxide, via stoichiometry, are found to be:

$n_{CO_2}=45.0g CaCO_3*\frac{1molCaCO_3}{100gCaCO_3}*\frac{1molCO_2}{1molCaCO_3}=0.45molCO_2$

Finally, since STP conditions are referred to a temperature of 273.15K and 1 atm, the volume, by using the ideal gas equation result:

$V=\frac{nRT}{P}=\frac{0.45mol*0.082\frac{atm*L}{mol*K}*273.15K }{1atm} \\ \\V=10.1L$

So the answer is B) 10.1 L.

Best regards.