How to solve mole problems?

Question

can anyone explain and show me steps for these problems?

thank you to all that try to answer.

2NaN3 (s) --> Na (s) + 3N2 (g)

How many grams of NaN3 are required to produce 10.0 square ft of nitrogen gas if the gas has a density of 1.25g/L?

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A piece of aluminum foil 0.550 mm thick and 1.00 cm square is allowed to react with bromine to form aluminum bromide.

How many moles of aluminum were used? The density of aluminum is 2.699 g/cm3)

How many grams of aluminum bromide form, assuming that the aluminum reacts completely?

Answer 1

Question 1 . It is impossible to produce square feet but cubic feet

1 cubic feet =28.31L so you must produce 283.2L. the mass of this is 283.2*1.25 =354g  so in moles of N2 354/28=12.67 moles of N2

you see that 3 moles of N2 correspond to 2 moles of NaN3 . So you need 12.67*2/3=8.42moles of NaN3  MW(23+3*14=65)

and you nees 8.42*65 =548g of NaN3

Question 2

mass of Aluminium V*d = 0.55*1*2.699=1.4845

in moles 1.4645/26.98=0.0542 mole of Al

AlBr3 molar mass =26.98+3*79.9=266.68

you multiply by 0.0542 =14.45g

Answer 2

2NaN3 (s) --> Na (s) + 3N2 (g)

How many grams of NaN3 are required to produce 10.0 cubic ft of nitrogen gas if the gas has a density of 1.25g/L?

since we have the density in litres, let's convert the 10.0 cu ft of N2 into litres:

10.0 cu ft @ 28.32  cu ft / litre = 283.2 liters

now use the density to get grams:

283. 2 litres @ 1.25 g/litre = 354  grams of nitrogen

now using the equation 2 NaN3 (@ 65.01 g/mol) --> 3 N2 (@28.01g/mol)

354 g N2 @ [(2)(65.01) grams NaN3 / (3) (28.01) grams N2] = 547.7 g NaN3

your answer is: 548 g NaN3

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A piece of aluminum foil 0.550 mm thick and 1.00 cm square is allowed to react with bromine to form aluminum bromide.

How many moles of aluminum were used? The density of aluminum is 2.699 g/cm3)

0.550 mm thick = 0.0550 cm thick

0.0550 cm thick times 1.00 cm square = 0.0550 cm3

0.0550 cm3 @ 2.699 g/cm3 = 0.148 grams of Al

0.148 g Al @ 26.98 g/mol =

your answer: 0.00550 moles of Al

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How many grams of aluminum bromide form, assuming that the aluminum reacts completely?

@  1 Al --> 1 AlBr3  :

0.00550 moles of Al  =  0.00550 moles of AlBr3

using its molar mass:

0.00550 moles of AlBr3 @ 133.34 g/mole =

your answer: 0.734 grams AlBr3 ...( aka  0.734 g of Al2Br6)