Question

If you weigh out a piece of Mg to a mass of 0.0240 g what is the minimum volume of 1.00 M HCl needed to ensure that the Mg is the limiting reactant?

Answer 1

1.97 ml.

Explanation

Mg reacts with HCl at a 1:2 ratio.

Mg + 2 HCl → MgCl₂ + H₂.

How many moles of atoms in 0.0240 g of Mg?

The molar mass $M$ of Mg is 24.305 g/mol. In other words, the mass of one mole of Mg atoms is 24.305 gram.

$n = \dfrac{m}{M} = \dfrac{0.0240}{24.305} = 9.87 \times 10^{-4} \; \text{mol}$.

How many moles of HCl does it take to oxidize that $9.87 \times 10^{-4} \; \text{mol}$ of Mg?

It takes 2 moles of HCl to oxidize 1 mole of Mg. As a result, it takes $2 \times 9.87 \times 10^{-4} = 1.97 \times 10^{-3}\; \text{mol}$ to oxidize $9.87 \times 10^{-4}\; \text{mol}$ of Mg.

Mg is the limiting reactant, meaning that HCl is in excess. There needs to be more than $1.97 \times 10^{-3}\; \text{mol}$ HCl for Mg to be the limiting reactant.

How many liters of 1.00 M HCl?

1.00 M is the same as 1.00 moles per liter.

$V = \dfrac{n}{c} = \dfrac{1.97 \times 10^{-3} \; \text{mol}}{1.00 \; \text{mol}\cdot \text{L}^{-1}} = 1.97 \times 10^{-3} \; \text{L} = 1.97 \; \text{mL}$.