Question

calculate the number of grams of methane required to produce 9 grams of water after combustion​

Answer 1

CH4 + 2O2 ---------> 2H2O + CO2

Given : 9g ot H2O

Thus moles of H2O;

• n = W/M
• n = 9/18 = 1/2 mole

Since O2 is on excess (present in air) thus CH4 is the limiting reagent.

Thus by stoichiometry;

1mol. CH4 = 2mol. H2O ....... (1)

By experimentally;

x mol. CH4 = 1/2 mol. H2O .........(2)

Thus by using unitary method in (1) & (2);

• 1 ------> 2
• x --------> 1/2

Therefore the value of x ;

• 2x = 1/2
• x = 1/4 moles of CH4

Given ;

• moles of CH4 = 1/4
• M(CH4) = 12 + 4 = 16 g/mol.

Thus for given mass;

• n = W/M
• W = M×n
• W = 16 × 1/4
• W = 4g

Thus we require 4g of CH4

#answerwithquality

#BAL