Wait...there is more! Tutor Partner has more to offer. Starting today, I'm adding problems on Chemistry here on my blog. The answer is provided for every problem. So stop working on those problems like you're doing homeworks all over again. Take a break; and then take your pick. Stoichiometry Problems 1. Based on the following equation: Na2CO3 + Ba(OH)2 --> BaCO3 + 2NaOH, calculate the number of mol of Na2CO3 required to yield 3.95 mol of NaOH. Answer: 1.98 mol of Na2CO3 2. Based on the following equation: 4FeS2 + 11O2 --> 2Fe2O3 + 8SO2, how many mol of O2 will be needed to produce 7.29 mol of SO2? Answer: 10.02 mol of O2 3. According to the equation: 4Fe + 3O2 --> 2Fe2O3, calculate the mass of O2 required so that 7.16 mol of Fe2O3 will be produced. Answer: 343.67 g of O2 4. According to the equation: 4Fe + 3O2 --> 2Fe2O3, how many g of Fe will be needed to produce 1.75 mol of Fe2O3? Answer: 195.46 g of Fe 5. Given the equation: 4Fe + 3O2 --> 2Fe2O3, how many mol of O2 is needed to produce 8.9 g of Fe2O3? Answer: 0.08 mol of O2 6. For the equation: NaHCO3 + HCl --> NaCl + CO2 + H2O, how many g of NaHCO3 is required so that 1.99 g of NaCl is produced? Answer: 2.86 g of NaHCO3 7. Based on the following equation: CaCO3 --> CaO + CO2, calculate the weight of CaCO3 necessary to produce 7.96 g of CO2. Answer:18.1 g of CaCO3 8. Based on the following equation: 4KClO3 --> 3KClO4 + KCl, calculate the amount of KClO4 that can be produced from the decomposition of 1.68 g of KClO3. Answer: 1.42 g of KClO4 9. According to the equation: 2KClO3 --> 2KCl + 3O2, calculate the number of mol of KClO3 required to yield 5.25 mol of O2. Answer: 3.5 mol of KClO3 10. Based on the following equation: Ni + 4CO --> Ni(CO)4, how many g of Ni will be needed to produce 4.34 mol of Ni(CO)4? Answer: 254.76 g of Ni Percent Composition Problems Calculate the percent composition of the following compounds: 1. PbBr2 Solution: Pb = 56.46% Br = 43.54% 2. FeCl2.4H2O Solution: H = 4.06% O = 32.19% Fe = 28.09% Cl = 35.66% 3. CuSO4 Solution: Cu = 39.81% S = 20.09% O = 40.1% 4. PbC2O4 Solution: Pb = 70.19% C = 8.14% O = 21.68% 5. MgCO3 Solution: Mg = 28.83% C = 14.25% O = 56.93% Limiting Reagent Problems 1. Based on the following equation: Ni + 4CO --> Ni(CO)4, calculate the weight of Ni(CO)4 which can be produced by the reaction of 1.81 g of Ni with 5.94 g of CO. Answer: Ni is the limiting reagent; 5.26 g of Ni(CO)4 can be produced. The weight of excess CO is 2.49 g. 2. Based on the following equation: Cu2S + O2 --> 2Cu + SO2, calculate the weight of Cu which can be produced by the reaction of 4.7 g of Cu2S with 1.83 g of O2. Answer: Cu2S is the limiting reagent; 3.75 g of Cu can be produced. The weight of excess O2 is 0.89 g. 3. Given the balanced equation: CaO + 2HCl --> CaCl2 + H2O, calculate the weight of H2O which can be produced by the reaction of 4.26 g of CaO with 4.8 g of HCl. Answer: HCl is the limiting reagent; 1.19 g of H2O can be produced. The weight of excess CaO is 0.57 g. 4. For the reaction: Ca(HCO3)2 + 2HCl --> CaCl2 + 2CO2 + 2H2O, calculate the weight of H2O which can be produced by the reaction of 9.92 g of Ca(HCO3)2 with 8.44 g of HCl. Answer: Ca(HCO3)2 is the limiting reagent; 2.2 g of H2O can be produced. The weight of excess HCl is 3.98 g. 5. According to the equation: CaO + 2HCl --> CaCl2 + H2O, calculate the weight of CaCl2 which can be produced by the reaction of 3 g of CaO with 6.02 g of HCl. Answer: CaO is the limiting reagent; 5.94 g of CaCl2 can be produced. The weight of excess HCl is 2.12 g.