Bài 2:
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
____0,2_____0,3_______0,1________0,3 (mol)
a, \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
d, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
___________0,3____0,3 (mol)
\(\Rightarrow m_{Cu}=0,3.64=19,2\left(g\right)\)
c2: \(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0.2 0.3 0.1 0.3 (mol)
\(m_{Al_2\left(SO_4\right)_3}=0.1\times342=34.2\left(g\right)\)
\(V_{H_2}=0.3\times22.4=6.72\left(l\right)\)
\(C_{MH_2SO_4}=\dfrac{0.3}{0.2}=1.5M\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0.3 0.3 (mol)
\(m_{Cu}=0.3\times64=19.2\left(g\right)\)
