\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
Mg + 2HCl ---> MgCl2 + H2
Fe + 2HCl ---> FeCl2 + H2
Theo các pthh: \(n_{HCl}=2n_{H_2}=2.0,3=0,6\left(mol\right)\)
Áp dụng ĐLBTKL:
mkim loại + mHCl = mmuối + mH2
=> mmuối = 13,6 + 0,6.36,5 - 0,3.2 = 34,9 (g)
=> D
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Fe}=y\end{matrix}\right.\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x x x ( mol )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+56y=13,6\\x+y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m=0,1.95+0,2.127=34,9g\)
=> Chọn D
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\Rightarrow n_{HCl}=2n_{H_2}=0,6mol\)
BTKL: \(m_A+m_{HCl}=m_{muối}+m_{H_2}\)
\(\Rightarrow13,6+0,6\cdot36,5=m_{muối}+0,3\cdot2\)
\(\Rightarrow m_{muối}=m=34,9g\)
Chọn D
Mg+Hcl->MgCl2+H2
x----------------------x
Fe+Hcl->Fecl2+H2
y---------------------y
=>\(\left\{{}\begin{matrix}24x+56y=13,6\\x+y=0,3\end{matrix}\right.\)
=>x=0,1 mol, y=0,2 mol
=>m =0,1.95+0,2.127=34,9g
=>d
nH2 = 6,72 : 22,4 = 0,3 (mol)
gọi số mol của Mg :a
số mol Fe : b
=> 24a + 56b= 13,6 (g)
pthh : Fe + 2HCl -> FeCl2 + H2
b b
Mg + 2HCl -> MgCl2 + H2
a a
=> a+b = 0,3
\(\left\{{}\begin{matrix}24a+56b=13,6\\a+b=0,3\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
pthh
: Fe +HCl -> FeCl2 + H2
0,2 0,2
Mg + 2HCl -> MgCl2 + H2
0,1 0,1
=> \(m_{hh}=\left(0,1.95\right)+\left(0,2.127\right)=34,9\left(G\right)\)
=> chọn D