Mole Ratio Worksheet

1. Let’s consider the equation CH4 (g) + H2O(g) ⇋ 3H2 (g) + CO(g) Write the molar ratios for a CH4/ H2O 1:1 =1 b H2O/ H2 1:3 = 0.33 c H2/ CH4 3:1 =3 I. II. III. If you use 1 mole of CH4 how many moles of H2 could be produced? Based on the mole ratio, 3:1, you could make 3 moles of H2 from 1 mole of CH4 If 7 moles of H2 were produced, how many moles of H2O would be required? First we determine the mole ratio (remember unknown species on top), H2O: H2, it is 1/3 or 0.33 Number of moles of given species x mole ratio = number of moles required of unknown. 7 x 0.33 = 2.33 moles of H2O needed. If 62.0g of CO were produced, how many moles of CH4 would have been required? First we need to convert the given value into moles 𝑚 So MCO = 12+ 16 = 28g/mol. Remember that n= 𝑀𝑤 Therefore 62/48 =2.214 moles Then we determine the mole ratio (remember unknown species on top) CH4: CO = 1:1, therefore 1x 2.214 = 2.21 moles of CH4 required. 2. Let’s consider the equation 8 H2(g) + S8(g) ⇋ 8H2S(g) Write the molar rations for a. b. c. I. H2/S8 8:1 = 8 H2/H2S 8:8 =1 S8/H2S 1:8 = 0.125 If you had 3.5 moles of H2 available and excess S8 what is the most number of moles of H2S you could produce? II. III. First we determine the mole ratio (remember unknown species on top), H2S: H2 = 8:8 =1 Number of moles of given species x mole ratio = number of moles required of unknown. So 3.5 x 1 = 3.5 moles of H2S could be produced. What would be the mass of this maximum yield? To answer this we need to convert our mole value for H2S into a mass value. Remember that m = nMw We can determine Mw = (2 x 1.008) + 32.06 = 34.08g/mol Therefore m = 3.5 x 34.08 = 119.3 grams of H2S could be produced. If you wanted to produce 3.00kg of H2S what mass of S8 would be required? Wow! This is as hard as these questions get! But don’t panic, you can do all of the steps already you just need to be systematic! First we need to convert the given value into moles. Remember that n= 𝑚 𝑀𝑤 So from above we know for H2S Mw = (2 x 1.008) + 32.06 = 34.08 g/mol m = 3.00kg = 3000g, therefore 3000/34.08 = 88.03moles of H2S we want to produce. Next we look at the question and it is asking what mass of S8 would be required? Given we now know the number of moles of H2S being produced we can use mole ratios to determine the number of moles of H2 needed. So (remember unknown species on top) S8: H2S = 1:8 = 0.125 Number of moles of given species x mole ratio = number of moles required of unknown. So, 0.125 x 88.03 = 11.0 mols of S8 gas Finally, the question is asking us to determine the mass. So remembering that m = nMw, We determine that Mw = (8 x 32.06) =256.48 g/mol therefore m = 11 x 256.48 = 2821 g or 2.82kg of S8 gas 3. Let’s consider the equation 2NH3(g) + 3CuO(s) ⇋ N2(g) + 3Cu(s) + 3H2O(g) Write the molar ratios for NH3/ CuO 2:3 = 0.66 CuO/ NH3 3:2 = 1.5 Cu/ H2O 3: 3 =1 NH3/Cu 2:3 = 0.66 Cu/ CuO 3:3 = 1 I. II. III. If you had 7 moles of CuO how many moles of NH3 would you need to react it all? First we determine the mole ratio (remember unknown species on top), NH3: CuO = 2:3 = 0.66 Number of moles of given species x mole ratio = number of moles required of unknown. So, 7 x 0.66 = 4.66 moles of NH3 needed If you needed to produce 6 moles of Cu(s) how many moles of NH3 would you need to react? First we determine the mole ratio (remember unknown species on top), NH3:Cu = 2:3 =0.66 Number of moles of given species x mole ratio = number of moles required of unknown. So, 6 x 0.66 = 4 moles of NH3 needed If you have 700.0kg of NH3 what mass of N2 will you produce? First we need to convert the given value into moles. Remember that n= 𝑚 𝑀𝑤 We can work out Mw NH3 = 14.01 + (3 x 1.008) = 17.034 g/mol m = 700.0kg = 700,000g, therefore700, 000/17.034 = 41094 moles of NH3 to react. Next we look at the question and it is asking what mass of N2 would be produced? Given we now know the number of moles of NH3 reacting we can use mole ratios to determine the number of moles of N2 produced. So (remember unknown species on top) N2: NH3 = 1:2 = 0.5 Number of moles of given species x mole ratio = number of moles required of unknown. So, 41094 x 0.5 = 20547 moles of N2 gas Finally, the question is asking us to determine the mass. So remembering that m = nMw, We determine that Mw = 2x 14.01= 28.02 g/mol therefore m = 20547 x 28.02 = 575731g = 575.7kg of N2 gas produced.