Chemistry help..please im stuck

Question

1) In the following equation, what volume of hydrogen gas is produced when 135 grams of aluminum are completely reacted with excess sulfuric acid?

2 Al 3 H2SO4 ---------------------------> 3 H2 Al2(SO4)3

2) In the reaction below, how many liters of ammonia gas are produced when 35 grams of liquid nitrogen completely react with excess hydrogen?

N 2 3H2 ----------------------> 2 NH3

Answer 1

1)

2 mol Al produces 3 mol H2

The molar mass of Al is 27.0 g/mol.

135 g Al x (1 mol / 27.0 g ) = 5.00 mol Al

5.00 mol Al x (3 mol H2 / 2 mol Al) = 7.50 mol H2

At STP, one mol of H2 occupies 22.4 L

7.50 mol H2 x (22.4 L / 1 mol ) = 168 L

2)

1 mol N2 produces 2 mol NH3

The molar mass of N2 is 28.0 g/mol.

35.0 g N2 x (1 mol / 28.0 g) = 1.25 mol N2

1.25 mol N2 x (2 mol NH3 / 1 mol N2) = 2.50 mol NH3

At STP, one mole of NH3 occupies 22.4 L

2.50 mol NH3 x (22.4 L / 1 mol ) = 56.0 L NH3

Answer 2

for both problems...steps are...

1) calculate moles aluminum

2) convert to moles of H2 via the balanced equation

3) calculate L of H2 via Pv = nRT

******** 1 ********

moles Al = mass / mw = 135 g / (27.0 g/mole) = 5 moles Al

from balanced equation 2 moles Al ---> 3 moles H2

so that...

5 moles Al x (3 moles H2/ 2 moles Al) = 7.5 moles H2

assuming STP (plug in other values if you have them)

PV = nRT

V = nRT / P = 7.5 moles x (.0821 L atm/moleK) x (273.15K) / (1 atm)

V = 168L

******** 2 ********

moles N2 = mass / mw = 35 g / (28.0 g/mole) = 1.25 moles N2

from balanced equation 1 moles N2 ---> 2 moles NH3

so that...

1.25 moles N2 x (2 moles NH3/ 1 moles N2) = 2.5 moles NH3

assuming STP (plug in other values if you have them)

PV = nRT

V = nRT / P = 2.5 moles x (.0821 L atm/moleK) x (273.15K) / (1 atm)

V = 56.1L

Answer 3

1)  First you have to figure the number of moles of Al = 135/26.96=5.0

     Then recognize that for every 2 mols of Al you produce 3 mols of H2

    So....5 mols al * 3 mols H2/2 mols Al = 7.5 mols H2 gas

     Finally, each mol of an ideal gas at standard temp. and pressure = 22.414 liters, so......7.5 mols * 22.414= 168.1 liters

2) As before, calculate mols of N2 gas :  35 g * 1mol/28 g = 1.25 mols

    for every mol of N2 gas, 2 mols of ammonia are formed, therefore

    2.5 mols of ammonia

    again each mol =22.414 liters therefore 22.414*2.5=56.0 liters