Đáp án:
Giải thích các bước giải:
Bài 1:
`Fe+2HCl → FeCl_2+H_2 ↑`
`n_{Fe}=\frac{2,8}{56}=0,05\ mol`
a) `n_{H_2}=n_{Fe}=0,05\ mol`
`V_{H_2}=0,05.22,4=1,12\ lít`
b) `n_{HCl}=2n_{Fe}=0,1\ mol`
`m_{HCl}=0,1.36,5=3,65\ gam`
Bài 2:
a) `2Al+6HCl→2AlCl_3+3H_2↑`
b) `n_{Al}=\frac{5,4}{27}=0,2\ mol`
`n_{HCl}=3n_{Al}=0,6\ mol`
`m_{HCl}=0,6.36,5=21,9\ gam`
c) `n_{H_2}=3/2 n_{Al}=0,3\ mol`
`V_{H_2}=0,3.22,4=6,72\ lít`
Bài 3:
\(CaCO_3 \xrightarrow{t^{0}} CaO+CO_2↓\)
a) `n_{CaO}=\frac{11,2}{56}=0,2\ mol`
`n_{CaCO_3}=n_{CaO}=0,2\ mol`
b) `n_{CaO}=\frac{7}{56}=0,125\ mol`
`n_{CaCO_3}=n_{CaO}=0,125\ mol`
`m_{CaCO_3}=0,125.100=12,5\ gam`
