\(CT:C_xH_4\)
\(\overline{M}=17\cdot2=34\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow12x+4=34\)
\(\Rightarrow x=2.5\)
\(n_{CO_2}=2.5\cdot n_{hh}=0.1\cdot2.5=0.25\left(mol\right)\)
\(m_{CO_2}=0.25\cdot44=11\left(g\right)\)
\(n_{H_2O}=2n_{hh}=2\cdot0.1=0.2\left(mol\right)\)
\(m_{\text{bình tăng}}=m_{CO_2}+m_{H_2O}=4.4+0.2\cdot18=8\left(g\right)\)
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