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