Question

Magnesium carbonate reacts with hydrochloric acid according to the equation:
MgCO3(s) + 2HCl(aq)  MgCl2(aq) + H2O(l) + CO2(g)
What mass of pure hydrochloric acid is needed to react with 4.2g of magnesium carbonate?

Answer 1

1 mole mgco3 require 2 mole HCl so 1mole mgco3 contain 84 grams so 84g react with 73 g HCl so 4.2 grams require 3.65 grams HCl

Answer 2

Answer : The mass of HCl acid needed is, 3.65 g

Solution : Given,

Mass of magnesium carbonate = 4.2 g

Molar mass of magnesium carbonate = 84 g/mole

Molar mass of hydrochloric acid = 36.5 g/mole

First we have to calculate the moles of magnesium carbonate.

$\text{Moles of }MgCO_3=\frac{\text{Mass of }MgCO_3}{\text{Molar mass of }MgCO_3}=\frac{4.2g}{84g/mole}=0.05mole$

Now we have to calculate the moles of HCl.

The balanced chemical reaction is,

$MgCO_3(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2O(l)+CO_2(g)$

From the balanced reaction, we conclude that

As, 1 mole of magnesium carbonate react with 2 moles of HCl

So, 0.05 mole of magnesium carbonate react with $0.05\times 2=0.1$ moles of HCl

Now we have to calculate the mass of HCl.

$\text{Mass of HCl}=\text{Moles of HCl}\times \text{Molar mass of HCl}=0.1mole\times 36.5g/mole=3.65g$

Therefore, the mass of pure HCl acid is, 3.65 g