12.00 g of hydrogen gas is mixed with 40.00 g of nitrogen gas. Assuming the reaction goes to completion, determine the maximum mass (in g) of ammonia that is produced. Give the answer to three significant figures.N2(g) + 3H2(g) ⇌ 2NH3(g)MN = 14.01 g mol–1MH = 1.01 g mol–1

First, we need to calculate the number of moles for both hydrogen and nitrogen.

For hydrogen: Number of moles = mass / molar mass = 12.00 g / 1.01 g mol–1 = 11.88 mol

For nitrogen: Number of moles = mass / molar mass = 40.00 g / 14.01 g mol–1 = 2.855 mol

From the balanced chemical equation, we can see that the molar ratio of nitrogen to hydrogen to ammonia is 1:3:2. This means that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Therefore, the limiting reactant is nitrogen because it needs 3 times the amount of hydrogen to react completely, but we only have about 4 times the amount of hydrogen.

So, the maximum number of moles of ammonia that can be produced is based on the amount of nitrogen.

Number of moles of ammonia = 2 * number of moles of nitrogen = 2 * 2.855 mol = 5.71 mol

Finally, we convert this back to grams using the molar mass of ammonia (which is 14.01 g/mol for nitrogen and 1.01 g/mol for hydrogen, times 3 for the three hydrogen atoms, times 2 for the two ammonia molecules).

Mass of ammonia = number of moles * molar mass = 5.71 mol * (2*(14.01 g/mol + 3*1.01 g/mol)) = 5.71 mol * 34.06 g/mol = 194.4626 g

So, the maximum mass of ammonia that can be produced is 194 g to three significant figures.

First, we need to calculate the number of moles for both hydrogen and nitrogen.

For hydrogen: Number of moles = mass / molar mass = 12.00 g / 1.01 g mol–1 = 11.88 mol

For nitrogen: Number of moles = mass / molar mass = 40.00 g / 14.01 g mol–1 = 2.855 mol

From the balanced chemical equation, we can see that the molar ratio of N2 to NH3 is 1:2 and the molar ratio of H2 to NH3 is 3:2.

Therefore, the number of moles of NH3 that can be produced from N2 is: = 2.855 mol * 2 = 5.71 mol

And the number of moles of NH3 that can be produced from H2 is: = 11.88 mol * (2