\(PTHH:\)
\(Mg+2HCl--->MgCl_2+H_2\)
\(nMg=\dfrac{2,4}{24}=0,1(mol)\)
\(mHCl=\dfrac{C\%HCl.mddHCl}{100}=\dfrac{10.36,5}{100}=3,65\left(g\right)\)
\(nHCl=\dfrac{3,65}{36,5}=0,1(mol)\)
So sánh: \(\dfrac{nH_2}{1}=0,1>\dfrac{nHCl}{2}=0,05\)
=> Mg còn dư sau phản ứng, chọn nHCl để tính
Theo PTHH: \(nH_2=\dfrac{1}{2}.nHCl=0,05(mol)\)
\(=> mH_2=0,05.2=0,1(g)\)
\(m dd sau = mMg+mddHCl-mH_2\)
\(=> m dd sau = 2,4+36,5-0,1=38,8(g)\)
Dung dich sau phản ứng là MgCl2
Theo PTHH: \(nMgCl_2=\dfrac{1}{2}.nHCl=0,05(mol)\)
\(=> mMgCl_2=0,05.95=4,75(g)\)
\(C\%MgCl_2=\dfrac{4,75.100}{38,8}=12,24\%\)
Ta có: \(C_M=\dfrac{10.C\%.D}{M}\)
\(<=> C_MMgCl_2=\dfrac{10.12,24.0,188}{95}=0,24M\)
