Asked by Kelsey
Determine the mass of carbon monoxide that is produced when 45.6 g of methane, ch4 react with 73.2 g of oxygen gas, o2. The products are carbon monoxide and water vapour
Answers
Answered by
DrBob222
This is a limiting reagent problem. You know that because BOTH reactants are given with their masses.
1. Write a balanced equation.
2CH4 + 3O2 --> 2CO + 4H2O
2. Convert 45.6 g CH4 and 73.2 g O2 to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles CO.
3b. Same procedure convert moles O2 to moles CO.
3c. It is more than likely that the answers frmo 3a and 3b will not be the same and one of them is wrong. The correct answer is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4. Convert moles from 3c to grams.
g = mole x molar mass.
1. Write a balanced equation.
2CH4 + 3O2 --> 2CO + 4H2O
2. Convert 45.6 g CH4 and 73.2 g O2 to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles CO.
3b. Same procedure convert moles O2 to moles CO.
3c. It is more than likely that the answers frmo 3a and 3b will not be the same and one of them is wrong. The correct answer is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
4. Convert moles from 3c to grams.
g = mole x molar mass.
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