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Zinaida [17]
3 years ago
10

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2

HCl ( aq ) ⟶ CaCl 2 ( aq ) + H 2 O ( l ) + CO 2 ( g ) CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) How many grams of calcium chloride will be produced when 32.0 g 32.0 g of calcium carbonate is combined with 10.0 g 10.0 g of hydrochloric acid?
Chemistry
1 answer:
e-lub [12.9K]3 years ago
4 0

<u>Answer:</u> The mass of calcium chloride formed is 15.21 grams.

<u>Explanation:</u>

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

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

  • <u>For calcium carbonate:</u>

Given mass of calcium carbonate = 32.0 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

\text{Moles of calcium carbonate}=\frac{32.0g}{100g/mol}=0.32mol

  • <u>For hydrochloric acid:</u>

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol

The given chemical equation follows:

CaCO_3(s)+2HCl(aq.)\rightarrow CaCl_2(aq.)+H_2O(l)+CO_2(g)

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium carbonate

So, 0.274 moles of hydrochloric acid will react with = \frac{1}{2}\times 0.274=0.137mol of calcium carbonate

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

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

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.274 moles of hydrochloric acid will produce = \frac{1}{2}\times 0.274=0.137moles of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 111 g/mol

Moles of calcium chloride = 0.137 moles

Putting values in equation 1, we get:

0.137mol=\frac{\text{Mass of calcium chloride}}{111g/mol}\\\\\text{Mass of calcium chloride}=(0.137mol\times 111g/mol)=15.21g

Hence, the mass of calcium chloride formed is 15.21 grams.

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