Cho 52g Kẽm tác dụng với 500g dung dịch Hcl(vừa đủ).Tính C%
2 câu trả lời
$n_{Zn}=\dfrac{52}{65}=0,8(mol)\\ PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Zn}=1,6(mol)\\ \Rightarrow m_{CT_{HCl}}=1,6.36,5=58,4(g)\\ \Rightarrow C\%_{HCl}=\dfrac{58,4}{500}.100\%=11,68\%$
$n_{Zn}=\dfrac{52}{65}=0,8(mol)$
PTHH:
$Zn+2HCl→ZnCl_2+H_2$
0,8 1,6
$m_{HCl}=1,6\times 36,5=58,4(gam)$
$C\text{%}$$=\dfrac{58,4}{500}.100=11,68$$\text{%}$