Bài 6:
a) \(m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)
b) \(n_{Cl_2}=\dfrac{\text{Số phân tử}}{6.10^{^{23}}}=\dfrac{3.10^{^{23}}}{6.10^{^{23}}}=0,5\left(mol\right)\)
`=>` \(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)
c) \(n_{CH_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
=>`` \(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)
d) \(m_{C_{12}H_{22}O_{11}}=1,5.342=513\left(g\right)\)
Bài 7:
a) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)
`=>` \(V_{CO_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)
b) \(n_{H_2S}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
`=>` \(V_{H_2S\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)
c) \(V_{Cl_2\left(đktc\right)}=n.22,4=0,7.22,4=15,68\left(l\right)\)
d) N phân tử H2 có 6.1023 phân tử H2
`=>` \(n_{H_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)
`=>` \(V_{H_2\left(đktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)
\(6a)m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)
\(b)n_{Cl_2}=\dfrac{\text{Số phân tử }Cl_2}{N}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
\(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)
\(c)n_{CH_4}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)
\(d)m_{C_6H_{12}O_6}=n.M=1,5.180=270\left(g\right)\)
\(7a)n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)
\(V_{CO_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
\(b)n_{H_2S}=\dfrac{\text{Số phân tử }H_2S}{N}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(V_{H_2S}=n.22,4=1,5.22,4=33,6\left(l\right)\)
\(c)V_{Cl_2}=n.22,4=0,7.22,4=15,68\left(l\right)\)
\(d)n_{H_2}=\dfrac{\text{Số phân tử }H_2}{N}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)
\(V_{H_2}=n.22,4=1.22,4=22,4\left(l\right)\)
