1)The reaction of methane and oxygen yields carbon dioxide gas and water. The unbalanced reaction is as follows: CH4(g) + O2(g)CO2(g) + H2O(g)

A. If 0.718 grams of methane are reacted, how many grams of water vapor are produced?

B. If 1.621 grams CO2 are released, how many grams of methane were reacted?

2)Write a balanced equation for the following reduction-oxidation reaction.

SO3–2 + MnO4–SO42– + Mn2+

1A).

Place an arrow where it belongs in the equation. Then balance it.
2. Convert grams you are given to moles. # moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles of what you were given to moles of water.
4. Convert moles water to grams. g = moles x molar mass.

1B. Use the same kind of procedure. This will work most stoichiometry problems.

#2. How much do you know about balancing redox equations. Here are a couple of hint to get you started.
S changes from +4 on the left to +6 on the right. Mn changes from +7 on the left to +2 on the right. You can make the problem simpler by using an arrow where it belongs. You do an arrow with --- and > or ===>.
Post your work if you get stuck.

i don't know if this is right, 2SO3–2 + MnO4 yield 4SO42– + 2Mn2+

sienes

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).