Bạn sửa đề thành 7.84 (l) nha
nH2 = 7.84/22.4 = 0.35 (mol)
Đặt :
nFe = x mol
nMg = y mol
mhh = 56x + 24y = 14.8 g (1)
Fe + H2SO4 => FeSO4 + H2
Mg + H2SO4 => MgSO4 + H2
nH2 = x + y = 0.35 (2)
(1) , (2) :
x = 0.2 => mFe = 0.2*56 = 11.2 g
y = 0.15 => mMg = 0.15*24=3.6 g
%Fe = 75.67%
%Mg = 24.33%
Gọi nFe = a (mol); nMg = b (mol)
=> 56a + 24b = 14,8
\(n_{H_2}=\dfrac{7,89}{22,4}=\dfrac{789}{2240}\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
______a-------------------------->a _________(Mol)
Mg + 2HCl --> MgCl2 + H2
_b--------------------------->b_______(mol)
=> \(\left\{{}\begin{matrix}56a+24b=14,8\\a+b=\dfrac{789}{2240}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=\dfrac{1777}{8960}\left(mol\right)\\b=\dfrac{197}{1280}\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{\dfrac{1777}{8960}.56}{14,8}.100\%=75\%\\\%m_{Mg}=\dfrac{\dfrac{197}{1280}.24}{14,8}.100\%=25\%\end{matrix}\right.\)
