Question

How much calcium carbonate in grams can be dissolved by 5.30 g of hcl? ( hint: begin by writing a balanced equation for the reaction between hydrochloric acid and calcium carbonate.)?

Answer 1

Answer: The mass of
`CaCO_3` dissolve can be, 7.25 grams.

Explanation : Given,

Mass of
`HCl` = 5.30 g

Molar mass of
`HCl` = 36.5 g/mol

Molar mass of
`CaCO_3` = 100 g/mol

First we have to calculate the moles of
`HCl`.

`\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}`

`\text{Moles of }HCl=(5.30g)/(36.5g/mol)=0.145mol`

Now we have to calculate the moles of
`CaCO_3`

The balanced chemical equation is:

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

From the reaction, we conclude that

As, 2 mole of
`HCl` react with 1 mole of
`CaCO_3`

So, 0.145 mole of
`HCl` react with
`(0.145)/(2)=0.0725` mole of
`CaCO_3`

Now we have to calculate the mass of
`CaCO_3`

`\text{ Mass of }CaCO_3=\text{ Moles of }CaCO_3* \text{ Molar mass of }CaCO_3`

`\text{ Mass of }CaCO_3=(0.0725moles)* (100g/mole)=7.25g`

Therefore, the mass of
`CaCO_3` dissolve can be, 7.25 grams.