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.

Step-by-step 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 = (m)/(M) = (0.0240)/(24.305) = 9.87 * 10^(-4) \; \text{mol}`.

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

It takes 2 moles of HCl to oxidize 1 mole of Mg. As a result, it takes
`2 * 9.87 * 10^(-4) = 1.97 * 10^(-3)\; \text{mol}` to oxidize
`9.87 * 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 * 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 = (n)/(c) = \frac{1.97 * 10^(-3) \; \text{mol}}{1.00 \; \text{mol}\cdot \text{L}^(-1)} = 1.97 * 10^(-3) \; \text{L} = 1.97 \; \text{mL}`.