Question

Mg metal reacts with HCl to produce hydrogen gas. Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)......A)What volume of hydrogen at 0 ∘C and 1.00 atm (STP) is released when 9.25 g of Mg reacts B)How many grams of magnesium are needed to prepare 3.90 L of H2 at 730 mmHg and 20 ∘C

Answer 1

A)

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

moles of Mg = 9.25 / 24 = 0.3854

moles of Mg = moles of H2

P V = n R T

1 x V = 0.385 x 0.0821 x 273

V = 8.64 L

volume of H2 gas = 8.64 L                                                  

B)

volume of H2 = 3.90 L

pressure = 730 mmHg = 0.961 atm

temperature = 293 K

P V = n R T

0.961 x 3.90 = n x 0.0821 x 293

n = 0.156

moles of H2 gas = 0.156

moles of Mg = 0.156

mass of Magnesium needed = 0.156 x 24

= 3.74 g