Đáp án:
Giải thích các bước giải:
\(\begin{array}{l} \text{Bài 1:}\\ PTHH:2H_2+O_2\xrightarrow{t^o} 2H_2O\\ n_{H_2O}=\dfrac{1,8}{18}=0,1\ mol.\\ Theo\ pt:\ n_{H_2}=n_{H_2O}=0,1\ mol.\\ ⇒V_{H_2}=0,1\times 22,4=2,24\ lít.\\ Theo\ pt:\ n_{O_2}=\dfrac12n_{H_2O}=0,05\ mol.\\ ⇒V_{O_2}=0,05\times 22,4=1,12\ lít.\\ \text{Bài 2:}\\ PTHH:BaO+H_2O\to Ba(OH)_2\\ Theo\ pt:\ n_{Ba(OH)_2}=n_{BaO}=2\ mol.\\ ⇒m_{Ba(OH)_2}=2\times 171=342\ g.\\ ⇒\text{Số phân tử}\ =2\times 6.10^{23}=12.10^{23}=1,2.10^{24}\\ \text{Bài 3:}\\ PTHH:P_2O_5+3H_2O\to 2H_3PO_4\\ \text{Sản phẩm là axit photphoric (H$_3$PO$_4$)}\\ n_{P_2O_5}=\dfrac{71}{142}=0,5\ mol.\\ Theo\ pt:\ n_{H_3PO_4}=2n_{P_2O_5}=1\ mol.\\ ⇒m_{H_3PO_4}=1\times 98=98\ g.\\ \text{Bài 4:}\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2↑\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ Theo\ pt:\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\ mol.\\ ⇒m_{Al}=0,1\times 27=2,7\ g.\\ \text{Bài 5:}\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2↑\\ n_{Fe}=\dfrac{1,12}{56}=0,02\ mol.\\ Theo\ pt:\ n_{FeSO_4}=n_{H_2}=n_{Fe}=0,02\ mol.\\ ⇒m_{FeSO_4}=0,02\times 152=3,04\ g.\\ V_{H_2}=0,02\times 22,4=0,448\ lít.\end{array}\)
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