Given the data n2(g)+o2(g)2no(g)+o2(g)2n2o(g)→→→2n o(g)2no2(g)2n2(g)+o2(g)δh=+180.7kjδ h=−113.1kjδh=−163.2kj use hess's law to calculate δh for the reaction n2o(g)+no2(g)→3no(g)

i dont know the answe

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

the answer is symmetrical : )