\(Mg\left(0,3\right)+H_2SO_4\left(0,3\right)\rightarrow MgSO_4\left(0,3\right)+H_2\left(0,3\right)\)
\(n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow a=0,3.24=7,2\left(g\right)\)
\(TheoPTHH:n_{H_2SO_4}=0,3\left(mol\right)\)\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(\Rightarrow b=\dfrac{29,4.100}{20}=147\left(g\right)\)
\(TheoPTHH:n_{MgSO_4}=0,3\left(mol\right)\)
\(mddsau =7,2+147-0,3.2=153,6(g)\)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,3.120.100}{153,6}=23,44\%\)
a)PTHH: Mg + H2SO4 ----> MgSO4 + H2\(\uparrow\)
n\(H_2\) = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH: n\(Mg\) = n\(H_2\) = 0,3 (mol)
=> mMg = 0,3.24 = 7,2 (g)
=> a = 7,2 (g)
Theo PTHH: n\(H_2SO_4\) = n\(H_2\) = 0,3 (mol)
=> m\(H_2SO_4\) = 0,3.98 = 29,4 (g)
=> m\(ddH_2SO_4\) = \(\dfrac{29,4.100\%}{20\%}=147\left(g\right)\)
=> b = 147 (g)
b) m\(H_2\) = 0,3.2 = 0,6 (g)
mddsp/ứ = 7,2 + 147 - 0,6 = 153,6 (g)
Theo PTHH: n\(MgSO_4\) = n\(H_2\) = 0,3 (mol)
=> m\(MgSO_4\) = 0,3.120 = 36 (g)
=> C%\(MgSO_4\) = \(\dfrac{36}{153,6}.100\%\approx23,44\%\)
