CaC2+2H2O->Ca(OH)2+C2H2
0,1-----------------------0,1 mol
m tạp chất = 3,6g =>m CaC2=10-3,6=6,4g
n CaC2=\(\dfrac{6,4}{64}\)=0,1 mol
=>VC2H2=0,1.22,4=2,24l
\(CaC_2+H_2O\rightarrow Ca\left(OH\right)_2+C_2H_2\)
\(m_{CaC_2}=\dfrac{\left[10.\left(100-36\right)\right]}{100}=6,4g\)
\(n_{CaC_2}=\dfrac{6,4}{64}=0,1mol\)
V(C2H2)= \(0,1.22,4=2,24l\)