Bài 10:
a) \(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: \(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
\(n_{HNO_3}=0,2\left(mol\right)\Rightarrow m_{HNO_3}=0,2.63=12,6\left(g\right)\)
b) \(n_{NaNO_3}=0,2\left(mol\right)\Rightarrow m_{NaNO_3}=0,2.85=17\left(g\right)\)
Bài 11:
a) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(n_{HCl}=0,6\left(mol\right)\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b) \(n_{FeCl_3}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)