Question

96.1 g of solid iron reacts with 0.0500 L of water (density = 0.997 g/mL) to form iron(III)oxide and hydrogen gas. If the reaction takes place in a 10.0 L container at 400.0 K, what pressure in atm would be measured?

When 0.62 mols of aluminum metal react with 4.55 mols of HCl according to the reaction below what will be the final pressure (in atm) if the two substances react in a solid steel chamber with a volume of 6.61L at a temperature of 165.1K? You should assume that there are no other substances in the chamber at the start of the reaction a nd that the change in volume of liquids and gasses is negligible . 2 Al(s) +6 HCl (l) --> 2 AlCl₃ (s) + 3 H₂ (g)

Answer 1

1. 3 Fe + 4 H2O Fe3O4 + 4 H2

Mol of Fe = Wt of Fe / Molar Mass

= 96.1 g / 56 g/mol

= 1.7 mol

Mass of H2O = 50 mL * 0.997 g/ mL

= 49.85 g

Mol of H2O = 49.85 g / 18 g/mol

= 2.78 mol

Now,

3 mol Fe : 4 mol H2O

1.7 mol Fe : 2.27 mol H2O

Thus,

Fe is limiting reagent

Now,

3 mol Fe : 4 mol H2

1.7 mol Fe : 2.27 mol H2

Now,

V= 10 L

n = 2.27 mol

R = 0.0821 L atm/Kmol

T = 400 K

PV = nRT

P * 10 L= 2.27 mol *0.0821 Latm / mol K * 400 K

P = 7.5 atm

2. 2 mol Al : 6 mol HCl

0.62 mol Al : 1.86 mol HCl

Thus,

Al is limiting reagent

Now,

2 mol Al : 3 mol H2

0.62 mol Al : 0.93 mol H2

V= 6.61 L

n= 0.93 mol

R= 0.0821 Latm / mol K

T= 165.1 K

PV = nRT

P * 6.61 L = 0.93 mol * 0.0821 Latm/mol K * 165.1 K

P = 1.9 atm