If the measured gas pressure in a collection flask during a gas-forming reaction was determined to be 1.44 atm? if the measured gas pressure in a collection flask during a gas-forming reaction was determined to be 1.44 atm, and the atmospheric pressure that day was measured as 0.95 atm, what is the partial pressure of the gas produced by the reaction?

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2