Question

H2 and O, react to form H,O according to the equation: 2H,(g)+02(8) 2H,0(8). If 2.00 g of hydrogen gas and 20.0 g of oxygen gas are placed in a 100.0 L container at 37.0°C, a) What is the total pressure of the mixture of gases at the end of the reaction? (hint identify your limiting reactant, identify the gases in the mixture at the end of the reaction) b) If the temperature was raised to 47°C, what would the new pressure be in the same container? s.

Answer 1

moles H2 = 2.00 g/ 2.016 g/mol=0.992
moles O2 = 20.0 g / 31.998 g/mol= 0.625

the ratio between H2 and O2 is 2 : 1 so H2 is the limiting reactant

moles O2 required = 0.992/2=0.496
moles O2 in excess = 0.625 - 0.496=0.129

moles H2O produced = 0.992

moles gases after the reaction :
moles H2 =0
moles O2 = 0.129
moles H2O = 0.992

total = 0.129 + 0.992 =1.12
T = 37 + 273 = 310 K

p = nRT/V = 1.12 x 0.08206 x 310 / 100.0 L=0.285 atm

T = 47 + 273 =320 K
p = 1.12 x 0.08206 x 320 / 100.0 L =0.294 atm