Phosphine, ph3(g), decomposes according to the equation 4ph3(g) --> p4(g) + 6h2(g) the kinetics of the decomposition of phosphine at 950 k was followed by measuring the total pressure in the system as a function of time. the data to the right were obtained in a run where the reaction chamber contained only pure phosphine at the start of the reaction.

good luck with that. I thought I had it, but it was not right.

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