\(m_{H_2SO_4}=9,8\%.500=49\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Pt: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,2mol 0,5mol → 0,2mol → 0,2mol
Lập tỉ số: \(n_{Mg}\) : \(n_{H_2SO_4}=0,2< 0,5\)
\(\Rightarrow Mg\) hết; H2SO4 dư
\(n_{H_2SO_4\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)
\(m_{H_2SO_4\left(dư\right)}=0,3.98=29,4\left(g\right)\)
\(m_{MgSO_4}=0,2.120=24\left(g\right)\)
\(m_{spư}=\left(m_{Mg}+m_{H_2SO_4}\right)-m_{H_2}\)
= \(\left(4,8+500\right)-0,2.2=201,92\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{201,92}.100=14,56\%\)
\(C\%_{MgSO_4}=\dfrac{24}{201,92}.100=11,88\%\)
