Question

What mass of water (in units of grams) is produced by the reaction of 5.25g CH3OH with an excess of O2 when the percent yield is 41.4%?

2CH3OH (g) + 3O2 (g) -> 2CO2 (g) + 4H2O (l)

Answer 1

2CH3OH (g) + 3O2 (g) -> 2CO2 (g) + 4H2O (l)

from the balanced equation it is clear that 2 moles of CH3OH will give 4 moles of H2O

no of moles of CH3OH = weight of CH3OH / molar mass of CH3OH

= 5.25g / 32.042 g/mol = 0.1638 moles

if 2 moles of CH3OH is giving 4 moles of H2O

0.1638 moles will give 2x 0.1638 moles = 0.3277 moles of water

weight of the water = moles of water x molar mass of H2O = 0.3277moles x 18g/moles = 5.9 grams water

so theritical weight of water = 5.9 grams water

% of yield = actual weight / theritical weight x 100

41.4 % = Actual weight / 5.9 x(100)

actual weight = 244.26/100 = 2.44 grams

Answer 2

70000000000000000