B1:\(PTHH:\text{2C2H2+5O2}\rightarrow\text{2H2O+4CO2}\)
Ta có nC2H2=10/22,4=25/56(mol)
\(\text{nO2=20/22,4=50/56(mol)}\)
\(\text{nC2H2 dư=5/56(mol)}\)
\(1.\left\{{}\begin{matrix}\text{VCO2=5/7x22,4=16(l)}\\\text{VH2O=5/14x22,4=8(l)}\end{matrix}\right.\)
\(\text{2) %C2H2=5/56:(5/56+5/7)x100=11,11%}\)
=>%CO2=88,89%