To solve these problems, we need to balance the chemical equation first. Once we have a balanced equation, we can use stoichiometry to determine the quantities of substances involved in the reaction.
1) Balancing the equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Now, we can calculate the answers to the questions:
A. If 0.718 grams of methane are reacted, we need to find the number of moles of methane first. To do that, we use the molar mass of methane (16.04 g/mol).
Number of moles of methane = 0.718 g / 16.04 g/mol = 0.0448 mol
From the balanced equation, we see that for every 1 mole of methane, 2 moles of water are produced. Therefore, the number of moles of water vapor produced is:
Number of moles of water = 2 * 0.0448 mol = 0.0896 mol
Now, we can convert the number of moles of water vapor to grams using the molar mass of water (18.02 g/mol).
Mass of water = 0.0896 mol * 18.02 g/mol = 1.6152 g
Therefore, 0.718 grams of methane yields 1.6152 grams of water vapor.
B. If 1.621 grams of CO2 are released, we need to find the number of moles of CO2. To do that, we use the molar mass of CO2 (44.01 g/mol).
Number of moles of CO2 = 1.621 g / 44.01 g/mol = 0.0368 mol
From the balanced equation, we see that for every 1 mole of methane, 1 mole of CO2 is produced. Therefore, the number of moles of methane reacted is:
Number of moles of methane = 0.0368 mol
Now, we can convert the number of moles of methane to grams using the molar mass of methane (16.04 g/mol).
Mass of methane = 0.0368 mol * 16.04 g/mol = 0.5907 g
Therefore, 1.621 grams of CO2 were produced from 0.5907 grams of methane.
2) The balanced equation for the reduction-oxidation reaction is:
SO3–2 + 2MnO4– + 2H+ → SO42– + 2Mn2+ + H2O
The balanced equation shows the transfer of electrons between the reactants and products. In this reaction, sulfur changes oxidation number from +4 to +6 (oxidation), and manganese changes oxidation number from +7 to +2 (reduction).