giả sử có 100 gam dd H2SO4
$=> nH_2SO_4= 0.15 mol$
$ACO_3 + H_2SO_4 ---> ASO_4+ H_2O + CO_2$
0.15..............0.15...................0.15......................0.15
mdd sau pư là :
0.15 A + 9 + 100 - 0.15*44= 0.15A+ 102.4 (gam)
mct ASO_4 = 0.15A+ 14.4 gam
=>mdd$ ASO_4= \dfrac{15A+ 1440}{17}$
ta có :
mdd= $\dfrac{15A+ 1440}{17}= 0.15A+ 102.4$
=>A= 24 (Mg)