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) \(n_{FeCl_3}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
\(10.\)
\(n_{NaOH}=\dfrac{m}{M}=\dfrac{8}{40}=0,2mol\)
PTHH: NaOH + HNO3 \(\rightarrow\) NaNO3 + H2O
TL: 1 1 1 1
mol: 0,2 \(\rightarrow\) 0,2 \(\rightarrow\) 0,2
\(a.m_{HNO_3}=n.M=0,2.61=12,2g\)
\(b.m_{NaNO_3}=n.M=0,2.83=16,6g\)
\(11.\)
\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{16}{160}=0,1mol\)
PTHH: Fe2O3 + 6HCl \(\rightarrow\) 2FeCl3 + 3H2O
TL: 1 6 2 3
mol: 0,1 \(\rightarrow\) 0,6 \(\rightarrow\) 0,2
\(a.m_{HCl}=n.M=0,6.36,5=21,9g\)
\(b.m_{FeCl_3}=n.M=0,2.162,5=32,5g\)