CaO + 3C → CaC2 + CO
m(C) = 10.2 g
n = m/M
n(C)=10.212=0.85 moln(C) = \frac{10.2}{12} = 0.85 \;moln(C)=1210.2=0.85mol
n(CaC2)=13n(C)n(CaC_2) = \frac{1}{3}n(C)n(CaC2)=31n(C)
n(CaC2)=13×0.85=0.283 moln(CaC_2) = \frac{1}{3} \times 0.85 = 0.283 \;moln(CaC2)=31×0.85=0.283mol
M(CaC2) = 40 + 24 = 64 g/mol
m(CaC2)=0.283×64=18.11 gm(CaC_2) = 0.283 \times 64 = 18.11 \;gm(CaC2)=0.283×64=18.11g
Answer: 18.11 g
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