Question

If you weigh out a piece of Mg to a mass of 0.0170 g, what is the minimum volume of 1.00 M HCl needed to ensure that the Mg is the limiting reactant?

Enter your answer to three significant figures in units of mL.

Answer 1

Balanced chemical equation is:

Mg + 2 HCl ---> MgCl2 + H2

Molar mass of Mg = 24.31 g/mol

mass(Mg)= 0.017 g

use:

number of mol of Mg,

n = mass of Mg/molar mass of Mg

=(1.7*10^-2 g)/(24.31 g/mol)

= 6.993*10^-4 mol

According to balanced equation

mol of HCl reacted = (2/1)* moles of Mg

= (2/1)*6.993*10^-4

= 1.399*10^-3 mol

This is number of moles of HCl

use:

M = number of mol / volume in L

1.0 = 1.399*10^-3/ volume in L

volume = 0.001399 L

volume = 1.399 mL

Answer: 1.40 mL