a) \(m_S=\dfrac{64.150}{100}=96(g)\\→n_S=\dfrac{96}{32}=3(mol)\\m_{Al}=\dfrac{36.150}{100}=54(mol)\\→n_{Al}=\dfrac{54}{27}=2(mol)\)
\(⇒\) CTHH: \(Al_2S_3\)
b) PTHH: \(2Al+3S→Al_2S_3\)
c) \(n_{Al}=\dfrac{5,4}{27}=0,2(mol)\)
\(n_{S}=\dfrac{10}{32}=0,3125(mol)\)
Theo pt: \(\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1(mol)\)
\(\dfrac{n_S}{3}=\dfrac{0,3125}{3}=0,104(mol)\)
\(→\dfrac{n_{Al}}{2}<\dfrac{n_S}{3}\)
\(→S\) dư \(0,104-0,1=0,004(mol)\)
\(→m_{S\,\,dư}=0,004.32=0,128(g)\)
Theo pt ta có tỉ lệ số mol: \(2:3:1\)
\(→n_{Al_2S_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1(mol)\)
\(→m_{Al_2S_3}=0,1.150=15(g)\)
Vậy khối lượng \(Al_2S_3\) là \(15g\) và khối lượng \(S\) dư là \(0,128g\)
