Ammonia can also be synthesized by this reaction: 3h2 (g) + n2 (g) → 2nh3 (g) what maximum amount of ammonia in grams can be synthesized from 25.2 g of n2 and 8.42 g of h2? express your answer in grams to one decimal place.

30.6 g First, determine the molar masses of the reactants and products. Start by looking up the atomic weights of all involved elements. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Molar mass N2 = 2 * 14.0067 = 28.0134 g/mol Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Now determine how many moles of each reactant we have Moles nitrogen gas = 25.2 g / 28.0134 g/mol = 0.899569492 mol Moles hydrogen gas = 8.42 g / 2.01588 g/mol = 4.176835923 mol Determine the limiting reactant. The balanced equation shows that for every mole of nitrogen gas, we need 3 moles of hydrogen gas. Since we have over 4 moles of hydrogen gas and less than 1 mole of nitrogen gas, it's obvious that the limiting reactant we have is nitrogen gas. And for every mole of nitrogen gas used, we produce 2 moles of ammonia, so the expected yield is 2 * 0.899569492 mol = 1.799138983 mol. Now we just need to multiply the number of moles of ammonia by the molar mass of ammonia, so 1.799138983 mol * 17.03052 g/mol = 30.64027244 g Rounding to 1 decimal place gives 30.6 g