Question

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) MgCl2 (aq) + H2 (g)

Answer 1

V=4.8 L
c=5.0 mol/L
M(Mg)=24.3 g/mol

1) n(HCl)=cv

2) m(Mg)=M(Mg)n(HCl)/2

3) m(Mg)=M(Mg)cv/2
    m(Mg)=24.3*5*4.8/2=291.6 g

Answer 2

Answer:

291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.

Explanation:

$Mg (s) + 2HCl (aq) \rightarrow MgCl_2(aq)+H_2(g)$

Concentration of 4,8 L  HCl solution = 5.0 M

Moles of HCl in solution = n

$Molarity=\frac{Moles}{Volume (L)}$

$5.0 M=\frac{n}{4.8L}$

n = 24 moles

According to reaction , 2 moles of HCl reacts with 1 mole of magnesium metal.

Then 24 moles of HCl will react with :

$\frac{1}{2}\times 24 = 12 mol$ of magnesium

Mass of 12 moles of magnesium = 12 mol × 24.31 g/mol=291.72g

291.72 grams of magnesium metal will react completely with 4.8 liters of 5.0 M HCl.