Asked by george
The following equation represents the partial combustion of methane,
CH4.
2CH4(g) + 3O2(g) > 2CO(g) + 4H2O(g)
At constant temperature and pressure, what is the maximum volume of carbon monoxide that can be obtained from 3.39 x 102 L of methane and 1.70 x 102 L of oxygen?
CH4.
2CH4(g) + 3O2(g) > 2CO(g) + 4H2O(g)
At constant temperature and pressure, what is the maximum volume of carbon monoxide that can be obtained from 3.39 x 102 L of methane and 1.70 x 102 L of oxygen?
Answers
Answered by
DrBob222
This is a limiting reagent (LR) problem.
Determine the limiting reagent.
Use the coefficients in the balanced equation to convert mols of the LR to mols of the product.
Then convert mols of the product to grams by = mols x molar mass.
Determine the limiting reagent.
Use the coefficients in the balanced equation to convert mols of the LR to mols of the product.
Then convert mols of the product to grams by = mols x molar mass.
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