Đáp án:
Giải thích các bước giải:
Bài 1:
\(1)\ SO_2\ \text{(oxit axit)}\\ 2)\ CuO\ \text{(oxit bazơ)}\\ 3)\ N_2O_5\ \text{(oxit axit)}\\ 4)\ HNO_3\ \text{(axit)}\\ 5)\ KOH\ \text{(bazơ)}\\ 6)\ MgO\ \text{(oxit bazơ)}\\ 7)\ Cu(NO_3)_2\ \text{(muối)}\\ 8)\ Al_2O_3\ \text{oxit lưỡng tính}\\ 9)\ P_2O_5\ \text{(oxit axit)}\\ 10)\ Zn(OH)_2\ \text{(bazơ)}\\ 11)\ FeCl_3\ \text{(muối)}\\ 12)\ H_2CO_3\ \text{(axit)}\\ 13)\ H_3PO_4\ \text{axit}\\ 14)\ Fe_2O_3\ \text{(oxit bazơ)}\\ 15)\ Fe(NO_3)_2\ \text{(muối)}\)
Bài 2:
\(a,\ S+O_2\xrightarrow{t^o} SO_2\\ b,\ 3Fe+2O_2\xrightarrow{t^o} Fe_3O_4\\ c,\ Zn+2HCl\to ZnCl_2+H_2↑\\ d,\ 2H_2+O_2\xrightarrow{t^o} 2H_2O\\ e,\ 2Na+2H_2O\to 2NaOH+H_2↑\\ f,\ P_2O_5+3H_2O\to 2H_3PO_4\)
Bài 3:
\(a,\ PTHH:Fe+2HCl\to FeCl_2+H_2↑\\ b,\ n_{Fe}=\dfrac{2,8}{56}=0,05\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{Fe}=0,1\ mol.\\ ⇒V_{HCl}=\dfrac{0,1}{2}=0,05\ lít.\\ c,\ Theo\ pt:\ n_{H_2}=n_{Fe}=0,05\ mol.\\ ⇒V_{H_2}=0,05\times 22,4=1,12\ lít.\)
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