Question

When 1 g of Caco3, reacts with 50 ml of 0.1M
HCI, the volume of Co2, produced is​

Answer 1

Answer:

You need to start off by writing out a balanced chemical equation which will show you the relationship between CaCO3 and CO2:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

This equation tells you that for every mol of CaCO3 consumed you will be producing 1 mol of CO2. In other words there is a 1:1 relationship between CaCO3 and CO2.

The next step is to determine the amount of CaCO3 you have in moles, and since HCl is in excess, you do not need to worry about which one is the limiting reagent (CaCO3 is limiting since HCl is in excess). You can do that by using the molar mass of CaCO3, which is 100 g/mol:

mol CaCO3 = 50 g CaCO3 x 1 mol CaCO3/100 g CaCO3 = 0.5 mol CaCO3

We now can calculate the amount in moles of the produced CO2 by using the relationship between them from the balanced chemical equation:

0.5 mol CaCO3 x 1 mol CO2/1 mol CaCO3 = 0.5 mol CO2.

It might seem redundant doing this as it is a 1:1 relationship for this specific reaction, however it is a good habit to get into even with the simplest of problems.

Since CO2 is a gas, we can use the ideal gas equation (PV=nRT) to interconvert between pressure (P), volume (V), mol (n) and temperature (T). The question states that we are under STP conditions which means that the pressure is 1 atm and the temperature is 273.15 K. All the variables other than volume are known:

P x V = n x R x T

1 atm x V = 0.5 mol x 0.08206 (L.atm/mol.K) x 273.15

V = 11.2 L

You could have also solved the problem directly after you calculated the number of moles of CO2 you produced, as under STP conditions, 1 mol of an ideal gas occupies 22.4 L:

0.5 mol * 22.4 L/mol = 11.2 L.

Answer 2

$\large\underline{\underline{\sf Solution:}}$

$\large{\sf CaCO_3+2HCl→CaCl_2+CO_2+H_2O}$

Initial Moles = $\sf{\frac{1}{100}}$ $\sf{\frac{1}{200}}$

Limiting reagent is HCl

2 moles of HCl form = 1 mole of $\sf{CO_2}$

•°•

$\sf{\frac{1}{200}}$ mole of HCl form = $\sf{\frac{1}{2}×\frac{1}{200}}$

$\large\implies{\sf \frac{1}{400}moles}$

$\large\implies{\sf \frac{1}{400}×22.4}$

$\large\implies{\sf 0.056\:litres }$

$\large\implies{\sf 0.056×1000}$

$\large\implies{\sf 56ml }$

$\Large\underline{\underline{\sf Answer:}}$

•°• Volume of $\sf{CO_2}$ produced is 56ml.