Đáp án:
PTHH: $Zn + 2HCI -> ZnCI_2 + H_2↑$
a, Ta có :
$n_{Zn}$ = $\frac{6,5}{65}$ = $0,1 (mol) =>$ $n_{HCI}$= $0,2 (mol)$
=> $m_{ddHCI}$ = $\frac{0,2.36,5}{10,95}$ ≈ 66,67 (g)
b, Theo PTHH:
$n_{ZnCI_2}$= $n_{H_2} = n_{Zn} = 0,1 (mol)$
$\left \{ {{m_{nCL2}= 0,1 . 136 = 13,6 (g)} \atop {m_{H_2}= 0,1.2=0,2(g)}} \right.$
Mặt khác :
$m_{dd} = m_{Zn} + m_{ddHCI} - m_{H_2} = 72,97 (g)$
=> C%ZnCI2 = => $\frac{13,6}{72,97}$ . 100% ≈ 18,64%
HỌC TỐT!