a.b.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,15 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)
c.\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,2 < 0,15 ( mol )
0,15 0,15 0,15 ( mol )
\(m_A=m_{CuO\left(dư\right)}+m_{Cu}=\left[\left(0,2-0,15\right).80\right]+\left[0,15.64\right]=4+9,6=13,6g\)