\(Zn+2HCl→\:ZnCl_2+H_2\)
0,1 → 0,2
số mol của kẽm là: \(n_{Zn}=\dfrac{m_{Zn}}{M_{Zn}}=\dfrac{6,5}{65}=0,1\left(\text{mol}\right)\)
⇒ \(n_{HCl}=2n_{Zn}=2\cdot0,1=0,2\left(\text{mol}\right)\)
khối lượng HCl là: \(m_{HCl}=n_{HCl}\cdot M_{HCl}=0,2\cdot36,5=7,3\left(\text{g}\right)\)
khối lượng dung dịch HCl là:
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%⇒\:m_{dd}=\dfrac{m_{ct}\cdot100\%}{C\%}=\dfrac{7,3\cdot100\%}{14,6\%}=50\left(g\right)\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{14,6\%}=50\left(g\right)=x\)