- Gọi \(\left\{{}\begin{matrix}n_{K_2SO_4}=a\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=b\left(mol\right)\end{matrix}\right.\)
=> nO = 4a + 12b (mol)
nnguyên tử trong hỗn hợp = 7a + 17b (mol)
Ta có: \(\dfrac{n_O}{n_{nguyên.tử.trong.hỗn.hợp}}=\dfrac{4a+12b}{7a+17b}=\dfrac{36}{55}\)
=> 32a = 48b
=> a = 1,5b
\(\%K_2SO_4=\dfrac{174a}{174a+342b}.100\%=43,28\%\)
\(\%Al_2\left(SO_4\right)_3=\dfrac{342b}{174a+342b}.100\%=56,72\%\)
- mhh (bđ) = 174a + 342b = 603b (g)
PTHH: \(K_2SO_4+BaCl_2\rightarrow BaSO_4+2KCl\)
a-------------------->a
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\)
b--------------------------------->3b
=> \(m_{BaSO_4}=233a+699b\left(g\right)\)
Xét \(\dfrac{m_{BaSO_4}}{m_{hh\left(bđ\right)}}=\dfrac{233a+699b}{603b}=\dfrac{233}{134}\)
=> Lượng kết tủa nặng gấp \(\dfrac{233}{134}\) lần hh ban đầu
