Bài 8.
\(n_{H_2}=\dfrac{0,28}{22,4}=0,0125mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,025 0,025 0,0125 ( mol )
\(m_{Na}=0,025.23=0,575g\)
\(m_{NaOH}=0,025.40=1g\)
Bài 9.
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,3 0,6 0,3 0,3 ( mol )
\(m_{Zn}=0,3.65=19,5g\)
\(m_{HCl}=0,6.36,5=21,9g\)
Cách 1.
\(m_{ZnCl_2}=0,3.136=40,8g\)
Cách 2.
\(m_{H_2}=0,3.2=0,6g\)
Áp dụng ĐL BTKL, ta có:
\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)
\(\Rightarrow m_{ZnCl_2}=19,5+21,9-0,6=40,8g\)
\(pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,3 0,6 0,3 0,3
=> \(\left\{{}\begin{matrix}m_{Zn}=0,3.65=19,5\left(g\right)\\m_{HCl}=0,6.36,5=21,9\left(g\right)\\m_{ZnCl_2}=136.0,2=27,1\left(G\right)\left(C1\right)\end{matrix}\right.\)
