\(n_{CaC_2}=\dfrac{m_{CaC_2}}{M_{CaC_2}}=\dfrac{6,4.80\%}{64}=0,08mol\)
\(CaC_2+2H_2O\rightarrow C_2H_2+Ca\left(OH\right)_2\)
0,08 0,08 ( mol )
\(V_{C_2H_2}=n_{C_2H_2}.22,4=0,08.22,4=1,792l\)
=> Chọn C
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