\(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\\ n_{H_2SO_4}=\dfrac{11,76\%.250}{100}=0,294\left(mol\right)\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ Vì:\dfrac{0,15}{1}>\dfrac{0,294}{3}\\ \Rightarrow Al_2O_3dư\\ n_{Al_2O_3}=n_{Al_2\left(SO_4\right)_3}=\dfrac{0,294}{3}=0,098\left(mol\right)\\ m_{ddsau}=0,098.102+250=259,996\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,096.342}{259,996}.100\approx12,628\%\)
\(n_{Al2O3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
\(m_{ct}=\dfrac{11,76.250}{100}=29,4\left(g\right)\)
\(n_{H2SO4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
Pt : \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)
1 3 1 3
0,15 0,3 0,1
Lập tỉ số so sánh : \(\dfrac{0,15}{1}>\dfrac{0,3}{3}\)
⇒ Al2O3 dư . H2SO4 phản ứng hết
⇒ Tính toán dựa vào số mol của H2SO4
\(n_{Al2\left(SO4\right)3}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)
\(m_{ddspư}=15,3+250=265,3\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{34,2.100}{265,3}=12,9\)0/0
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