\(SO_3+H_2O-->H_2SO_4\left(1\right)\)
0,1________________0,1
a)
\(n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\)
=> nSO3 hết , H2O dư
=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
=> \(m_{d^2}=142+8=150\left(g\right)\)
=> \(C\%_{H_2SO_4}=\dfrac{9,8}{150}.100=6,5\%\)
b) \(CuO+H_2SO_4-->CuSO_4+H_2O\left(2\right)\)
0,1____ 0,1______________0,1
=> \(n_{CuO}=n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)
=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)
=> \(m_{CuO}=0,1.80=8\left(g\right)\)
=>\(m_{d_2sau}=150+8=158\left(g\right)\)
=> \(C\%_{CuSO_4}=\dfrac{16}{158}.100=10,13\%\)
c)
\(Mg+CuSO_4-->MgSO_4+Cu\left(3\right)\)
0,1_____0,1_________0,1
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
=> Mg hết
=> \(m_{d^2sau}=4,8+158-0,1.64-0,1.24=154\left(g\right)\)
=> \(C\%_{MgSO_4}=\dfrac{0,1.120}{154}.100=7,79\left(g\right)\)