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 dа ta:

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 %