Question

What is the mass of HCl that would be required to completely react with 5.2 grams of Mg

Answer 1

Balanced chemical equation :

2 HCl + Mg = H2 + MgCl2

2 x 36.5 g ------------- 24.30 g Mg
 Mass HCl g ----------------- 5.2 g Mg

5.2 x 2 x 36.5 / 24.30 =

Therefore:

379.6 / 24.30 => 15.62 g of HCl 

Answer 2

Answer : The mass of HCl required will be, 16.06 grams

Explanation : Given,

Mass of Mg = 5.2 g

Molar mass of Mg = 24 g/mole

Molar mass of HCl = 36.5 g/mole

First we have to calculate the moles of Mg.

$\text{Moles of }Mg=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{5.2g}{24g/mole}=0.22moles$

Now we have to calculate the moles of HCl.

The given balanced reaction is,

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

From the balanced reaction we conclude that

As, 1 mole of Mg react with 2 mole of HCl

So, 0.22 moles of Mg react with $0.22\times 2=0.44$ moles of HCl

Now we have to calculate the mass of HCl.

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(0.44mole)\times (36.5g/mole)=16.06g$

Therefore, the mass of HCl required will be, 16.06 grams