Molar mass of Na2SO4,
MM = 2*MM(Na) + 1*MM(S) + 4*MM(O)
= 2*22.99 + 1*32.07 + 4*16.0
= 142.05 g/mol
mass(Na2SO4)= 15.0 g
use:
number of mol of Na2SO4,
n = mass of Na2SO4/molar mass of Na2SO4
=(15 g)/(1.421*10^2 g/mol)
= 0.1056 mol
Molar mass of C = 12.01 g/mol
mass(C)= 17.5 g
use:
number of mol of C,
n = mass of C/molar mass of C
=(17.5 g)/(12.01 g/mol)
= 1.457 mol
Balanced chemical equation is:
Na2SO4 + 4 C ---> Na2S + 4 CO
1 mol of Na2SO4 reacts with 4 mol of C
for 0.1056 mol of Na2SO4, 0.4224 mol of C is required
But we have 1.457 mol of C
so, Na2SO4 is limiting reagent
we will use Na2SO4 in further calculation
Molar mass of Na2S,
MM = 2*MM(Na) + 1*MM(S)
= 2*22.99 + 1*32.07
= 78.05 g/mol
According to balanced equation
mol of Na2S formed = (1/1)* moles of Na2SO4
= (1/1)*0.1056
= 0.1056 mol
use:
mass of Na2S = number of mol * molar mass
= 0.1056*78.05
= 8.242 g
Answer: 8.24 g
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