Use the following balanced equation to answer the parts below:3H2+N2--> 2NH3How many grams of NH3 can be made from 2.76 mol of H2 and excess N2?How many grams of NH3 can be made from 1.45 mol of N2 and excess H2?If you want to make 24.83 mol of NH3, how many grams of N2 do you need?

It is clear from the given reaction that 3 mole of hydrogen gas reacts with one mole of nitrogen gas to form two mole of ammonia. The amount of nitrogen is in excess. So, it act as an excess reagent. Limiting reagent is the reagent which is consumed in the chemical reaction completely. It decides the amount of formation of desired product. Here limiting reagent is H2. Calculate the number of moles of ammonia formed from 2.76 mol of hydrogen gas. nNH3=2 mol3 mol×2.76 mol=1.84 mol Therefore, number of moles of NH3 is 1.84 mol. Molar mass of NH3 is 17 g/mol. Calculate the mass of NH3. mNH3=17 g/mol×1.84 mol=31.28 g Therefore, mass of ammonia is 31.28 g.The amount of hydrogen is in excess. So, it act as an excess reagent. Limiting reagent is the reagent which is consumed in the chemical reaction completely. It decides the amount of formation of desired product. Here limiting reagent is N2. Calculate the number of moles of ammonia formed from 1.45 mol of nitrogen gas. nNH3=2 mol1 mol×1.45 mol=2.9 mol Therefore, number of moles of NH3 is 2.9 mol. Molar mass of NH3is17 g/mol. Calculate the mass of NH3. mNH3=17 g/mol×2.9 mol=49.3 g Therefore, mass of ammonia is 49.3 g.Number of moles of ammonia is 24.83 mol. Calculate the number of moles of nitrogen formed from 24.83 mol of ammonia. nN2=1 mol2 mol×24.83 mol=12.415 mol Therefore, number of moles of N2 is 12.415 mol. Molar mass of N2 is 28 g/mol. Calculate the mass of N2. mN2=28 g/mol×12.415 mol=347.62 g Therefore, mass of nitrogen gas N2 is 347.62 g.