Question

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Answer 1

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 

b) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\)

c) Theo PTHH: \(n_{H_2}=\dfrac{0,1\cdot3}{2}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2,25^o1bar}=0,15\cdot24,79=3,7185\left(l\right)\)