A) To determine how many moles of NH3 react with 5.64 mol of O2, we need to use the stoichiometric ratio between NH3 and O2 in the balanced equation.
From the balanced equation, we can see that the ratio between NH3 and O2 is 4:7. This means that for every 4 moles of NH3, we need 7 moles of O2.
Using this ratio, we can set up a proportion:
(4 moles NH3) / (7 moles O2) = (x moles NH3) / (5.64 moles O2)
Cross-multiplying and solving for x gives:
(4 moles NH3) * (5.64 moles O2) = (7 moles O2) * (x moles NH3)
x = (4 moles NH3 * 5.64 moles O2) / 7 moles O2
x = 3.22 moles NH3
Therefore, 3.22 moles of NH3 will react with 5.64 mol of O2.
B) To determine how many moles of NO2 are obtained from 3.27 mol of O2, we again use the stoichiometric ratio between O2 and NO2 from the balanced equation.
From the balanced equation, we can see that the ratio between O2 and NO2 is 7:4.
Using this ratio, we can set up a proportion:
(7 moles O2) / (4 moles NO2) = (3.27 moles O2) / (x moles NO2)
Cross-multiplying and solving for x gives:
(7 moles O2) * (x moles NO2) = (4 moles NO2) * (3.27 moles O2)
x = (7 moles O2 * 3.27 moles O2) / 4 moles NO2
x = 5.71 moles NO2
Therefore, 5.71 moles of NO2 are obtained from 3.27 mol of O2.
C) To determine how many moles of H2O will be produced from 8.95 g of NH3, we need to convert grams of NH3 to moles using the molar mass of NH3.
The molar mass of NH3 is calculated as follows:
(Molar mass of N) + 3 * (Molar mass of H)
= (14.01 g/mol) + 3 * (1.01 g/mol)
= 17.04 g/mol
Now, we can set up a conversion:
(8.95 g NH3) * (1 mol NH3 / 17.04 g NH3) * (6 mol H2O / 4 mol NH3)
Simplifying:
(8.95 / 17.04) * (6 / 4) = 2.37 moles H2O
Therefore, 2.37 moles of H2O will be produced from 8.95 g of NH3.
D) To determine how many grams of NH3 will be needed to produce 0.0160 g NO2, we need to use the stoichiometric ratio between NH3 and NO2 from the balanced equation.
From the balanced equation, we can see that the ratio between NH3 and NO2 is 4:4, meaning the ratio is 1:1.
Given that the mass of NO2 is 0.0160 g, the mass of NH3 needed can be found using the molar mass of NH3.
The molar mass of NH3 is 17.04 g/mol. So, we can set up the following conversion:
(0.0160 g NO2) * (1 mol NO2 / 46.0 g NO2) * (1 mol NH3 / 1 mol NO2) * (17.04 g NH3 / 1 mol NH3)
Simplifying:
(0.0160 / 46.0) * (17.04 / 1) = 0.00591 g NH3
Therefore, 0.00591 g of NH3 will be needed to produce 0.0160 g of NO2.