\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
0,1 0,1 ( mol )
\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
\(m_{ddspứ}=10,2+300=310,2\left(g\right)\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{310,2}.100=11,02\%\)
a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
b) $n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$
Theo PTHH :
$n_{H_2SO_4} = 3n_{Al_2O_3} = 0,3(mol)$
$\Rightarrow m_{dd\ sau\ pư} = 300 + 10,2 = 310,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,1.342}{310,2}.100\% = 11\%$
a) $Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O$
b) $n_{Al_2O_3} = \dfrac{10,2}{102} = 0,1(mol)$
Theo PTHH :
$n_{H_2SO_4} = 3n_{Al_2O_3} = 0,3(mol)$
Gọi $m_{dd\ H_2SO_4} = a(gam)$
$\Rightarrow m_{dd\ sau\ pư} = 300 + 10,2 = 310,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,1.342}{310,2}.100\% = 11\%$
