Question

Limestone is a sedimentary rock composed largely of the mineral calcite, (calcium carbonate) CaCO3. To test the purity of the calcium carbonate, a sample of the mineral is reacted with hydrochloric acid to produce calcium chloride, carbon dioxide and water If a 15.70 g sample of calcite is reacted with excess hydrochloric acid and 16.89 g of calcium chloride are produced, what is the percent by mass of the calcium carbonate in the limestone? A) 69.4% B)93.0 % C) 96.94% D)98,22 %

Answer 1

`C\text{ : 96.94 \%}`

Step-by-step explanation:

Here, we want to get the percentage by mass of calcium carbonate in the limestone

We start by writing an equation of chemical reaction

The reaction involves calcium carbonate reacting with hydrochloric acid to give calcium chloride, carbon dioxide and water

We have the equation as follows:

`\text{CaCO}_{3(s)\text{ }}+2HCl_((aq))\rightarrow CaCl_{2(aq)\text{ }}+CO_{2(g)\text{ }}+H_2O_((l))`

Now, let us work with the extra information

16.89 g of calcium chloride are produced

Let us get the number of moles of calcium chloride produced

We can get that by dividing the mass by the molar mass of calcium chloride

The molar mass of calcium chloride is 111 g/mol

So, the number of moles will be:

`(16.89)/(111)\text{ = 0.1522 mole}`

From the equation,

1 mole of calcite gave 1 mole of calcium chloride, then 0.1522 mole of calcium chloride was produced by 0.1522 mole of calcite

To get the mass of calcite that produced this, we have to multipply the number of moles by the molar mass of calcite

The molar mass of calcite is 100 g/mol

That means the mass of calcite that produced the calcium chloride will be:

`0.1522\text{ }*\text{ 100 = 15.22 g}`

What this mean is that 15.22 g of calcite produced the given mass of calcium chloride

The percentage by mass of calcite in the limestone will be:

`(15.22)/(15.70)*\text{ 100\% = 96.94 \%}`