Calculating the Amount of Product Formed from a Limiting Reactant The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3): How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2? When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction 2 2 CO2(g)+4 H2O(g), what is the excess reactant and how many moles of it remains at the end of the reaction? (a) 9 mol CH3OH(/), (b) 10 mol CO2(g), (c) 10 mol CH3OH(/), (d) 14 mol CH3OH(l), (e) 1 mol O2(g).
Answer: In the formation of ammonia we need to observe that in the formation process we make nitrogen atom to get reacted with 3 moles of hydrogen to give 2 moles of ammonia . The question asked is how many moles of ammonia formed from 3 moles of nitrogen and 6 moles of hydrogen . Obviously when one mole of nitrogen and 3 moles of hydrogen on reacting gives the 2 moles of ammonia .then when 3 moles of nitrogen and 6 moles of hydrogen are used the we get double the amount of ammonia than we get in the habers process.so we get 4 moles of ammonia . Definition: The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated. If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants will be entirely consumed while another will be left over. The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant. How to Find the Limiting Reagent There are two ways to determine the limiting reagent. One method is to find and compare the mole ratio of the reactants used in the reaction (approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the smallest amount of product is the limiting reagent (approach 2). Approach 1: Find the limiting reagent by looking at the number of moles of each reactant. 1. Determine the balanced chemical equation for the chemical reaction. 2. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor). 3. Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio. 4. Use the amount of limiting reactant to calculate the amount of product produced. 5. If necessary, calculate how much is left in excess of the non-limiting reagent. Approach 2: Find the limiting reactant by calculating and comparing the amount of product each reactant will produce. 1. Balance the chemical equation for the chemical reaction. 2. Convert the given information into moles. 3. Use stoichiometry for each individual reactant to find the mass of product produced. 4. The reactant that produces a lesser amount of product is the limiting reagent. 5. The reactant that produces a larger amount of product is the excess reagent. 6. To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given. Consider the balanced chemical equation for the combustion reaction of methanol as follows: 2CH3OH (l)+3O2(g) 2CO2(g) +4H2O(g). For every 3 moles of oxygen, it requires 2 moles of methanol for the complete conversion of reactants into products. For 1 mole of oxygen, the number of moles of methanol required is as follows: Number of moles of methanol required= *1mol For 15 moles of oxygen, the number of moles of methanol required is as follows: Number of moles of methanol requried = *15mol =10 mol . There are 24 moles of methanol are available. Thus, the limiting reagent is oxygen. Then, the methanol is left in the reaction mixture. Calculate the number of moles of methanol left in the reaction mixture as follows: Number of moles of methanol left =24 mol-10 mol =14 mol Hence the correct option is (d)14 mol CH3OH.