cho 1.82g hh MgO và Al2O3 t/d vừa đủ với 250ml dd H2SO4 0,2M. a/viết pthh b/tính k/l muối tạo thành sau p/ư

2 câu trả lời

a, MgO + H2SO4 -> MgSO4 + H2O Al2O3 + 3H2SO4 -> Al2(SO4)3 + 3H2O b, nH2SO4=0,25*0,2=0,05 mol Nhận thấy nH2SO4=nH2O=0,05 mol mH2O=0,05*18=0,9g Áp dụng ĐLBTKL m muối=mhh + mH2SO4 - mH2O = 1,82 + 0,05*98 - 0,9 =5,82g

Em tham khảo nha :

\(\begin{array}{l}
a)\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
b)\\
{n_{{H_2}S{O_4}}} = 0,25 \times 0,2 = 0,05mol\\
{n_{{H_2}O}} = {n_{{H_2}S{O_4}}} = 0,05mol\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{hh}} + {m_{{H_2}S{O_4}}} = {m_m} + {m_{{H_2}O}}\\
 \Rightarrow {m_m} = 1,82 + 0,05 \times 98 - 0,05 \times 18 = 5,82g
\end{array}\)

 

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