Based on the equation below, how many grams of nitrogen gas (N2) will be produced from the decomposition of one mole of sodium azide (NaN3)? Use a molar mass of 28.0 grams for nitrogen gas (N2).

2NaN3 → 2Na + 3N2
A. 56.0 grams
B. 42.0 grams
C. 28.0 grams
D. 18.7 grams

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