Question

If 50g of CaCO3 is treated with 50g of hcl how many grams of co2 can be produced acc. To following equation CaCO3 +2hcl gives CaCl2 + CO2 + H20

Answer 1

Answer:

the co2 formed will be 22 gram

Answer 2

The mass of carbon dioxide produced is 22 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For calcium carbonate:

Given mass of calcium carbonate = 50 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{50g}{100g/mol}=0.5mol$

• For HCl:

Given mass of HCl = 50 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of HCl}=\frac{50g}{36.5g/mol}=1.37mol$

The chemical equation for the reaction of calcium carbonate and HCl follows:

$CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O$

By Stoichiometry of the reaction:

1 mole of calcium carbonate reacts with 2 moles of HCl

So, 0.5 moles of calcium carbonate will react with = $\frac{2}{1}\times 0.5=1mol$ of HCl

As, given amount of HCl is more than the required amount. So, it is considered as an excess reagent.

Thus, calcium carbonate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium carbonate produces 1 mole of carbon dioxide

So, 0.5 moles of calcium carbonate will produce = $\frac{1}{1}\times 0.5=0.5moles$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.5 moles

Putting values in equation 1, we get:

$0.5mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.5mol\times 44g/mol)=22g$

Learn more about limiting reagent:

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