Question

(a). How many grams of water will be produced from 14 g of methane?

CH4 (g) + O2 (g) --> CO2 (g) + H2O (l)

(b). How many grams of oxygen is needed for the full combustion of 14g of methane?

Answer 1

a)

first, convert 14 g of methane to moles

1 mol of methane, CH4 = 16 g

then x mol of methane = 14 g

calcualte x

x = 14 /16 mol = 0.875 mol of CH4

first, ensure the reaction is balanced:

CH4 (g) + O2 (g) --> CO2 (g) + H2O (l)

balance C, then H, finally O

CH4 (g) + O2 (g) --> CO2 (g) + 2H2O (l)

this is now balanced:

CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (l)

1 mol of methane = 2 mol of water

0.875 mol of CH4 = 2x0.875 = 1.75 mol of H2O expected

change to mass

1 mol o H2O = 18 g

1.75 mol of H2O = y g

y = 18*1.75 = 31.5 g of H2O expected

b)

now, calcualte O2 required...

from ratio

1 mol of CH4 = 2 mol of O2

0.875 mol of CH4 will need then 2x0.875 = 1.75 mol of O2

then

1 mol of O2 = 32 g

1.75 mol of O2 = 32x1.75 = 56 g of O2 will be required for complete combusiton