Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation
3H2(g)+N2(g)→2NH3(g) PArt 1: How many grams of NH3 can be produced from 3.26 mol of N2 and excess H2. Part 2:
How many grams of H2 are needed to produce 13.17 g of NH3? part 3 How many molecules (not moles) of NH3 are produced from 1.10×10−4 g of H2?
To answer these questions, we need to use stoichiometry, which is a mathematical relationship between the amounts of reactants and products in a chemical reaction.
Part 1: How many grams of NH3 can be produced from 3.26 mol of N2 and excess H2?
To determine the grams of NH3 produced, we need to use the balanced equation and the molar ratios. The coefficients in the balanced equation represent the ratios of moles between reactants and products.
From the balanced equation:
3H2(g) + N2(g) -> 2NH3(g)
The coefficient of N2 is 1, and the coefficient of NH3 is 2. This means that the molar ratio of N2 to NH3 is 1:2.
Given that we have 3.26 mol of N2, we can calculate the moles of NH3 produced using the molar ratio:
3.26 mol N2 * (2 mol NH3 / 1 mol N2) = 6.52 mol NH3
Now, to determine the grams of NH3 produced, we need to use the molar mass of NH3. The molar mass of NH3 is the sum of the atomic masses of its elements: 1(atomic mass of N) + 3(atomic mass of H) = 17.03 g/mol.
6.52 mol NH3 * 17.03 g/mol = 111.20 g NH3
Therefore, 3.26 mol of N2 can produce 111.20 g of NH3.
Part 2: How many grams of H2 are needed to produce 13.17 g of NH3?
In this case, we need to use the molar mass and the molar ratio between H2 and NH3.
Given that we have 13.17 g of NH3, we can first calculate the number of moles of NH3:
13.17 g NH3 * (1 mol NH3 / 17.03 g NH3) = 0.7735 mol NH3
Now, using the molar ratio from the balanced equation, the ratio of H2 to NH3 is 3:2:
0.7735 mol NH3 * (3 mol H2 / 2 mol NH3) = 1.160 mol H2
Finally, to determine the grams of H2 needed, we use the molar mass of H2, which is 2.02 g/mol:
1.160 mol H2 * 2.02 g/mol = 2.34 g H2
Therefore, 13.17 g of NH3 requires 2.34 g of H2.
Part 3: How many molecules of NH3 are produced from 1.10×10−4 g of H2?
In this case, we need to use the molar mass, Avogadro's number, and the molar ratio between H2 and NH3.
First, we need to convert the given mass of H2 to moles. The molar mass of H2 is 2.02 g/mol.
1.10×10−4 g H2 * (1 mol H2 / 2.02 g H2) = 5.45×10−5 mol H2
Now, using the molar ratio from the balanced equation (3:2), the ratio of H2 to NH3 is 3:2:
5.45×10−5 mol H2 * (2 mol NH3 / 3 mol H2) = 3.63×10−5 mol NH3
To convert from moles to molecules, we use Avogadro's number, which is 6.02×10^23 molecules/mol:
3.63×10−5 mol NH3 * 6.02×10^23 molecules/mol = 2.19×10^19 molecules NH3
Therefore, 1.10×10−4 g of H2 produces 2.19×10^19 molecules of NH3.