\(m_{CaCO_3}=80\%.120=96\left(g\right)\\ n_{CaCO_3}=\dfrac{96}{100}=0,96\left(mol\right)\\ CaCO_3\underrightarrow{to}CaO+CO_2\\ n_{CaO\left(LT\right)}=n_{CaCO_3}=0,96\left(mol\right)\\ n_{CaO\left(TT\right)}=90\%.n_{CaO\left(LT\right)}=90\%.0,96=0,864\left(mol\right)\\ \rightarrow m_{CaO\left(TT\right)}=0,864.56=48,384\left(g\right)\)
=> CHỌN B
Chọn B
Ta có
m CaCO3 = 120 . 80% = 96 ( g )
=> n CaCO3 = 0,96 ( mol )
=> n CaO = 0,96 ( mol ) => m CaO ( lí thuyết ) = 53,76 ( g )
=> m CaO ( thu được ) = 53,76 . 90% = 48,38 ( g )
\(m_{CaCO3}=\dfrac{120.80}{100}=96\left(g\right)\)
\(n_{CaCO3}=\dfrac{96}{100}=0,96\left(g\right)\)
Pt : \(CaCO_3\rightarrow\left(t_o\right)CaO+CO_2|\)
1 1 1
0,96 0,96
\(n_{CaO}=\dfrac{0,96.1}{1}=0,96\left(mol\right)\)
⇒ \(m_{CaO}=0,96.56=53,76\left(g\right)\)
\(m_{CaO_{\left(tt\right)}}=\dfrac{53,76.90}{100}\approx48,384\left(g\right)\)
⇒ Chọn câu : B
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