b/
\(2R+6H_2SO_4\rightarrow R_2\left(SO_4\right)_3+3SO_2+6H_2O\)
0,15 0,45 0,075 0,225 (mol)
n\(_{SO_2}=\)\(\dfrac{5,04}{22,4}\)=0,225(mol)
M\(_R=\dfrac{4,05}{0,15}=27\left(g\right)\)
\(\rightarrow\)nhôm(Al)
b/
m\(_{H_2SO_{4\left(cần\right)}}\)=\(\dfrac{0,45.98.100}{80}=55,125\left(g\right)\)
\(m_{H_2SO_{4\left(lấy\right)}}=55,125.105\%=57,88125\left(g\right)\)
m\(_{dd}\)=4,05+57,88125=61,93125(g)
m\(_{Al_2\left(SO_4\right)_3=0,075.342=25,65\left(g\right)}\)
C%\(_{Al_2\left(SO_4\right)_3}=\dfrac{25,65}{\text{61,93125}}.100\approx41,4\%\)
C%\(_{H_2SO_{4\left(dư\right)}}=\dfrac{55,125.5}{\text{61,93125}.100}.100\approx4,45\%\)