If 2 CH3OH + 3 O2 -> 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent yield be if the reaction started with 229 g of methanol in excess oxygen gas

Percent yield = 84.5 %

Explanation:

Given data:

Mass of methanol = 229 g

Actual yield of water = 219 g

Percent yield of water = ?

Solution:

Chemical equation:

2CH₃OH + 3O₂  →  2CO₂  + 4H₂O

Number of moles of methanol:

Number of moles = mass/ molar mass

Number of moles = 229 g/ 32 g/mol

Number of moles = 7.2 mol

Now we will compare the moles of water with methanol.

                        CH₃OH         :            H₂O

                            2               :               4

                           7.2             :           4/2×7.2 = 14.4 mol

Mass of water:

Mass = number of moles × molae mass

Mass = 14.4 mol × 18 g/mol

Mass = 259.2 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 219 g / 259.2 g × 100

Percent yield = 84.5 %

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