How many grams of magnesium metal will react completely with 4.8 liters of 5.0 m hcl? show all of the work needed to solve this problem. mg (s) + 2hcl (aq) yields mgcl2 (aq) + h2 (g)

291.6 g

Solution step-by-step:

1) Chemical equation

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

2) molar ratios

1 mol Mg (s) : 2 mol HCl (aq) : 1 mol MgCl2(aq) : 1 mol H2 (g)

3) Determine the number of moles of HCl in 4.8 liters of solution 5.0 M

M = n / V => n = M*V = 5.0M * 4.8 liters = 24 moles of HCl

4) Use the molar ratios between Mg and HCl to build a proportion and solve for the unnkown:

1 mol Mg              x
=  => x = 24 mol HCl * 1 mol Mg / 2 mol HCl
2 mol HCl       24 mol HCl

x = 12 mol Mg.

5) Convert 12 mol Mg to grams using the atomic mass of Mg

Aomtic mass of Mg = 24.3 g/mol

Mass in grams = number of moles * atomic mass = 12 mol * 24.3 g/mol = 291.6 g.

291.6 g

The correct answer would be 1. precipitation