5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is ignitted by an electric spark. if the reaction product is gaseous water and the temperature of the cylinder is maintained at 35Celsius what is the final gas pressure inside the cylinder?

For the limiting reagent 2.48mol of H2O

the excess 3.13mol of H2O ...

n=moles m=mass Mr=Relative molecular mass

n=m/MrsoH2: n1=5/2=40/16=2.5O2: n2=50/32=25/16=1.5625

2 H2 + O2 - > 2 H2O (oxygen is in execess) 2.5 1.5625 0-2.5 - 1.25 + 2.50 0.3125 2.5 there for 0.3125+2.5=2.8125 mole of gas in cylinder

P*V=n*R*TP=7.8925 atm

1) Balanced chemical reaction:

2H2 + O2 - > 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed - > 0 mol at the end

O2 = 1.25 mol consumed - > 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O - > 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT

n = 2.8125

V = 9 liters

R = 0.082 atm*lit/K*mol

T = 35 C + 273.15 = 308.15K

p = nRT/V = 7.9 atm