Câu hỏi của Muon Lam Quen

Question

cho 4,2 g hh(Mg,Zn) tac dung het HCl thay thoat ra 2,24 lit khi H2 (dktc) tinh m muoi khan

ai jup vs

Lê Thu Dương · Accepted Answer

$Mg+2HCl-->MgCl2+H2$

$Zn+2HCl-->ZnCl2+H2$

$n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)$

$m_{H2}=0,1.2=0,2\left(g\right)$

Theo pthh

$\sum n_{HCl}=2n_{H2}=0,2\left(mol\right)$

$m_{HCl}=0,2.36,5=7,3\left(g\right)$

Áp dụng ĐLBTKL

$m_{muối}=m_{KL}+m_{HCl}-m_{H2}$

=$4,2+7,3-0,2=11,3\left(g\right)$Câu trả lời: $Mg+2HCl-->MgCl2+H2$

$Zn+2HCl-->ZnCl2+H2$

$n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)$

$m_{H2}=0,1.2=0,2\left(g\right)$

Theo pthh

$\sum n_{HCl}=2n_{H2}=0,2\left(mol\right)$

$m_{HCl}=0,2.36,5=7,3\left(g\right)$

Áp dụng ĐLBTKL

$m_{muối}=m_{KL}+m_{HCl}-m_{H2}$

=$4,2+7,3-0,2=11,3\left(g\right)$

Tống Linh Trang · Answer

PTHH:

Mg+2HCl=>MgCl2+H2

Zn+2HCl=>ZnCl2+H2

nH2=2.24/22,4=0.1(mol)

mH2=0,1.2=0,2(g)

Theo PTHH:

∑nHCl = 2nH2 = 0, 2 (mol)

mHCl=0,2.36,5=7,3(g)

Aps dụng ĐLBTKL:

mmuối=mKL+mHCl-mH2

=4,2+7,3-0,2=11,3(g)

Câu trả lời: PTHH: