Answer:
1. The balanced equation is 2KCIO3 → 2KCI + 3O2. According to the law of conservation of mass, the mass of the reactants must equal the mass of the products. Therefore, the mass of oxygen produced is:
Mass of oxygen = Mass of KCIO3 - Mass of KCI
Mass of oxygen = 500 g - 303 g
Mass of oxygen = 197 g
Therefore, 197 g of O2 are produced.
2. The balanced equation is N2 + 3H2 → 2NH3. We need to find out how much H2 is needed to react with 100 g of N2 to produce 121 g of NH3. First, we need to calculate the number of moles of N2 and NH3:
Moles of N2 = Mass of N2 / Molar mass of N2
Moles of N2 = 100 g / 28 g/mol
Moles of N2 = 3.57 mol
Moles of NH3 = Mass of NH3 / Molar mass of NH3
Moles of NH3 = 121 g / 17 g/mol
Moles of NH3 = 7.12 mol
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the number of moles of H2 needed is:
Moles of H2 = Moles of N2 x (3/1)
Moles of H2 = 3.57 mol x 3
Moles of H2 = 10.71 mol
Finally, we can calculate the mass of H2 needed:
Mass of H2 = Moles of H2 x Molar mass of H2
Mass of H2 = 10.71 mol x 2 g/mol
Mass of H2 = 21.42 g
Therefore, 21.42 g of H2 are needed.
3. The balanced equation is 4Fe + 3O2 → 2Fe2O3. We need to find out how much oxygen is needed to react with 350 g of Fe to produce 500 g of Fe2O3. First, we need to calculate the number of moles of Fe and Fe2O3:
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 350 g / 55.85 g/mol
Moles of Fe = 6.26 mol
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
Moles of Fe2O3 = 500 g / 159.69 g/mol
Moles of Fe2O3 = 3.13 mol
According to the balanced equation, 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3. Therefore, the number of moles of O2 needed is:
Moles of O2 = Moles of Fe x (3/4)
Moles of O2 = 6.26 mol x (3/4)
Moles of O2 = 4.69 mol
Finally, we can calculate the mass of O2 needed:
Mass of O2 = Moles of O2 x Molar mass of O2
Mass of O2 = 4.69 mol x 32 g/mol
Mass of O2 = 150.08 g
Therefore, 150.08 g of O2 are needed.
4. The balanced equation is CH2 + 2O2 → CO2 + 2H2O. We know that 16 g of CH2 reacts with 64 g of O2 to produce 44 g of CO2. We need to find out how much water is produced. First, we need to calculate the number of moles of CH2 and CO2:
Moles of CH2 = Mass of CH2 / Molar mass of CH2
Moles of CH2 = 16 g / 14 g/mol
Moles of CH2 = 1.14 mol
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of CO2 = 44 g / 44 g/mol
Moles of CO2 = 1 mol
According to the balanced equation, 1 mole of CH2 reacts with 2 moles of O2 to produce 2 moles of H2O. Therefore, the number of moles of H2O produced is:
Moles of H2O = Moles of CH2 x (2/1)
Moles of H2O = 1.14 mol x 2
Moles of H2O = 2.28 mol
Finally, we can calculate the mass of H2O produced:
Mass of H2O = Moles of H2O x Molar mass of H2O
Mass of H2O = 2.28 mol x 18 g/mol
Mass of H2O = 41.04 g
Therefore, 41.04 g of H2O are produced.
5. The balanced equation is CaCO3 → CaO + CO2. We need to find out how much CO2 is produced from the decomposition of 200 g of CaCO3 if 112 g of CaO are produced. First, we need to calculate the number of moles of CaCO3 and CaO:
Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
Moles of CaCO3 = 200 g / 100.09 g/mol
Moles of CaCO3 = 1.999 mol
Moles of CaO = Mass of CaO / Molar mass of CaO
Moles of CaO = 112 g / 56.08 g/mol
Moles of CaO = 1.999 mol
According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CaO and 1 mole of CO2. Therefore, the number of moles of CO2 produced is:
Moles of CO2 = Moles of CaCO3 x (1/1)
Moles of CO2 = 1.999 mol
Finally, we can calculate the mass of CO2 produced:
Mass of CO2 = Moles of CO2 x Molar mass of CO2
Mass of CO2 = 1.999 mol x 44 g/mol
Mass of CO2 = 87.96 g
Therefore, 87.96 g of CO2 are produced.