Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced according to the balanced chemical equation below:

CaCO3(s) + 2HC1(aq) 4 CaC12(aq) +H20(1) + CO2(g)

Imagine mixing 2.5 moles of calcium carbonate with 4.8 moles of hydrochloric acid. Calculate the number of moles of CaCO3 and HCL used in the intro activity. determine the limiting reactant and use the ideal gas law to estimate the max volume of CO2

Answer 1

HCl is limiting reactant.

The maximum volume of the 2.4 moles of carbon dioxide at STP is 53.76 L.

Step-by-step explanation:

`CaCO_3(s) + 2HC1(aq) \rightarrow4 CaCl_2(aq) +H_2O(l) + CO_2(g)`

Moles of calcium carbonate = 2.5 moles (given)

According to reaction, 1 mole of calcium carbonate react with 2 moles of HCl. then 2.5 moles of calcium carbonate will react with:

`(2)/(1)* 2.5 mol=5.0 mol` of HCl

Moles of hydrochloric acid = 4.8 moles (given)

According to reaction, 2 mole of HCl react with 1 moles of calcium carbonate. Then 4.8 moles of HCl will react with:

`(1)/(2)* 4.8 mol=2.4 mol` of calcium carbonate

As we can see that HCl is completely reacting with 2.4 moles of calcium carbonate.Hence HCl is limiting reagent .Also, the moles of carbon dioxide gas will depend upon moles of HCl.

According to reaction, 2 moles of HCl gives 1 mole of carbon dioxide gas. then 4.8 moles of HCl will give:

`(1)/(2)* 4.8 mol=2.4 mol` of carbon dioxide gas.

To calculate volume of carbon dioxde gas at :

1 mole of ideal gas occupies 22.4 L of volume at STP.

Then 2.4 moles of carbon dioxide will occupy :

`2.4* 22.4 L=53.76 L`

The maximum volume of the 2.4 moles of carbon dioxide at STP is 53.76 L.