Hóa học cho 5g kem vao dung dich H2SO4 co 2 mol H2SO4, tinh khoi luong chat san pham? 04/10/2021 By Jasmine cho 5g kem vao dung dich H2SO4 co 2 mol H2SO4, tinh khoi luong chat san pham?
$PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$ $n_{Zn}=\dfrac{5}{65}=\dfrac{1}{13}mol.$ $⇒HCl$ $dư.$ $Theo$ $pt:$ $n_{ZnSO_4}=n_{H_2}=n_{Zn}=\dfrac{1}{13}mol.$ $⇒m_{ZnSO_4}=\dfrac{1}{13}.161=12,38g.$ $⇒m_{H_2}=\dfrac{1}{13}.2=0,154g.$ chúc bạn học tốt! Trả lời
$n_{Zn}=\frac{5}{65}=\frac{1}{13} mol$ $Zn+H_2SO_4\to ZnSO_4+H_2$ $\Rightarrow n_{ZnSO_4}= n_{H_2}=\frac{1}{13} mol$ (dư axit) $m_{ZnSO_4}=\frac{1}{13}.161=12,38g$ $m_{H_2}=\frac{1}{13}.2=0,154g$ Trả lời
$PTPƯ:Zn+H_2SO_4\xrightarrow{} ZnSO_4+H_2↑$
$n_{Zn}=\dfrac{5}{65}=\dfrac{1}{13}mol.$
$⇒HCl$ $dư.$
$Theo$ $pt:$ $n_{ZnSO_4}=n_{H_2}=n_{Zn}=\dfrac{1}{13}mol.$
$⇒m_{ZnSO_4}=\dfrac{1}{13}.161=12,38g.$
$⇒m_{H_2}=\dfrac{1}{13}.2=0,154g.$
chúc bạn học tốt!
$n_{Zn}=\frac{5}{65}=\frac{1}{13} mol$
$Zn+H_2SO_4\to ZnSO_4+H_2$
$\Rightarrow n_{ZnSO_4}= n_{H_2}=\frac{1}{13} mol$ (dư axit)
$m_{ZnSO_4}=\frac{1}{13}.161=12,38g$
$m_{H_2}=\frac{1}{13}.2=0,154g$