\(a)n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{Mg}=n_{MgSO_4}=n_{H_2}=0,3mol\\ x=m_{Mg}=0,3.24=7,2g\\ b)m_{MgSO_4}=0,3.120=36g\\ c)C_{M_{MgSO_4}}=\dfrac{0,3}{0,16}=1,875M\)
Đúng 2
Bình luận (0)
\(n_{H_2}=\dfrac{7.437}{22.4}=0.332\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(n_{Mg}=n_{MgSO_4}=n_{H_2}=0.332\left(mol\right)\)
\(m_{Mg}=x=0.332\cdot24=7.968\left(g\right)\)
\(m_{MgSO_4}=0.332\cdot120=39.84\left(g\right)\)
\(C_{M_{MgSO_4}}=\dfrac{0.332}{0.16}=2.075\left(M\right)\)
Đúng 0
Bình luận (2)
