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 …

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asked Apr 3, 2020 229k views

1 Answer

Kitti Wateesatogkij · Answer 1 · 2020-04-09T19:41:11+0000

Answer: The mass of calcium chloride formed is 15.21 grams.

Step-by-step explanation:

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

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

Given mass of calcium carbonate = 32.0 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=(32.0g)/(100g/mol)=0.32mol$

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=(10.0g)/(36.5g/mol)=0.274mol$

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 =
(1)/(2)* 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 =
(1)/(2)* 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:

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

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