a)
\(m_{dd.H_2SO_4.24,5\%}=\dfrac{58,8.100}{24,5}=240\left(tấn\right)\)
b)
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{200.24,5\%}{98}=0,5\left(mol\right)\)
PTHH: Mg + H2SO4 --> MgSO4 + H2
Xét tỉ lệ: \(\dfrac{0,5}{1}=\dfrac{0,5}{1}\) => Phản ứng vừa đủ
PTHH: Mg + H2SO4 --> MgSO4 + H2
0,5--------------->0,5-----0,5
\(m_{dd.sau.pư}=12+200-0,5.2=211\left(g\right)\)
\(C\%_{MgSO_4}=\dfrac{0,5.120}{211}.100\%=28,436\%\)