(1)Aluminum reacts with oxygen to produce aluminum oxide according to the following reaction:

4Al(s) + 3O2(g) → 2Al2O3(s)
Calculate the moles of Al2O3 produced when the reaction is performed with 31.06 g of each reactant.

(2)Nitrogen dioxide reacts with water to produce nitric acid and nitrogen monoxide according to the following reaction:
3NO2(g) + H2O(g) → 2HNO3(g) + NO(g)
Calculate the moles of nitric acid produced when 46.79 grams of each reactant is used.

(3)Ammonia reacts with oxygen to produce nitrogen monoxide and water according to the following equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Calculate the moles of water produced in this reaction when 88.82 grams of each reactant are used.

(4)A rocket can be powered by the reaction of dinitrogen tetroxide (N2O4) and hydrazine (N2H4):
N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)
A rocket was designed to hold 4.16 kilograms dinitrogen tetroxide and excess hydrazine. How much nitrogen gas in grams would be produced in this rocket?

#1, 2, and 3 are limiting reagent problems. #4 is a plain vanilla stoichiometry problem.

Here are worked examples for both. Just follow the steps.

http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html

http://www.jiskha.com/science/chemistry/stoichiometry.html

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To calculate the moles of a substance in a chemical reaction, you need to use the molar mass and the given mass of the substance. The molar mass is the mass of one mole of a substance and is expressed in grams/mole.

(1) Aluminum oxide (Al2O3) has a molar mass of 101.96 g/mol. To find the moles of Al2O3 produced, you can use the formula:

moles = mass (in grams) / molar mass

Given that 31.06 g of aluminum (Al) is reacted, the moles of Al can be calculated using the molar mass of aluminum (26.98 g/mol):

moles of Al = 31.06 g / 26.98 g/mol

Since the stoichiometry of the reaction is 4 moles of Al for every 2 moles of Al2O3 produced, you can use the stoichiometric ratio to find the moles of Al2O3:

moles of Al2O3 = (moles of Al / 4) * 2

(2) Nitric acid (HNO3) has a molar mass of 63.01 g/mol. To find the moles of HNO3 produced, you can use the formula:

moles = mass (in grams) / molar mass

Given that 46.79 g of nitrogen dioxide (NO2) is reacted, the moles of NO2 can be calculated using the molar mass of NO2 (46.01 g/mol):

moles of NO2 = 46.79 g / 46.01 g/mol

Since the stoichiometry of the reaction is 3 moles of NO2 for every 2 moles of HNO3 produced, you can use the stoichiometric ratio to find the moles of HNO3:

moles of HNO3 = (moles of NO2 / 3) * 2

(3) Water (H2O) has a molar mass of 18.02 g/mol. To find the moles of H2O produced, you can use the formula:

moles = mass (in grams) / molar mass

Given that 88.82 g of ammonia (NH3) and oxygen (O2) are reacted, the moles of NH3 and O2 can be calculated using their respective molar masses (17.03 g/mol for NH3 and 32.00 g/mol for O2):

moles of NH3 = 88.82 g / 17.03 g/mol
moles of O2 = 88.82 g / 32.00 g/mol

Since the stoichiometry of the reaction is 4 moles of NH3 for every 6 moles of H2O produced, you can use the stoichiometric ratio to find the moles of H2O:

moles of H2O = (moles of NH3 / 4) * 6

(4) To find the moles of nitrogen gas (N2) produced in this rocket, you need to convert the given mass of dinitrogen tetroxide (N2O4) into moles and then use the stoichiometry of the reaction.

Given that the rocket holds 4.16 kilograms (kg) of N2O4, you first need to convert this mass into grams:

mass of N2O4 = 4.16 kg * 1000 g/kg = 4160 g

Next, you can calculate the moles of N2O4 using its molar mass (92.02 g/mol):

moles of N2O4 = 4160 g / 92.02 g/mol

Since the stoichiometry of the reaction is 1 mole of N2O4 for every 3 moles of N2 produced, you can use the stoichiometric ratio to find the moles of N2:

moles of N2 = (moles of N2O4 / 1) * 3

Finally, to find the mass of N2 in grams, you multiply the moles of N2 by the molar mass of N2 (28.02 g/mol):

mass of N2 = moles of N2 * 28.02 g/mol