A 200 g bar of magnesium metal is placed in 2.40 L of 6.00 mol/L hydrochloric acid solution. What is the final mass of the magnesium bar?

Question

A 200 g bar of magnesium metal is placed in 2.40 L of 6.00 mol/L hydrochloric acid solution. What is the final mass of the magnesium bar?

Answer 1

First, write the correctly balanced equation for the reaction in question:

...Mg(s) + 2HCl(aq) ==> MgCl₂(aq) + H₂(g)

Next, find which reactant (Mg, or HCl) is in limiting supply:

...moles Mg = 200 g x 1 mole/24.3 g = 8.23 moles

...moles HCl = 6.00 mol/L x 2.40 L = 14.4 moles

...It takes 2 HCl : 1 Mg, so HCl is limiting b/c you don't have twice as much HCl as Mg.

Next, determine moles and mass of Mg that are used up in the reaction:

...moles Mg used up = 14.4 moles HCl x 1 mole Mg/2 mole HCl = 7.2 moles Mg used up

...mass of Mg used up = 7.2 moles Mg x 24.3 g/mol = 174.96 g Mg used up

Finally, calculate mass of Mg left over:

...200 g - 174.96 g = 25.04 g = 25 g (to 2 sig. figs.) or 30 g (to 1 sig. fig. which is what you have in 200 g)