Bài 1: cho các chất sau CO2, Fe, K2O, Al2O3, KOH a) chất nào tác dụng với H2O b) chất nào tác dụng với H2SO4 c) chất nào tác dụng NaOH. Viết pt minh họa Bài 2: Cho 3,2g S tác dụng với 4,48 lít O2 ở đktc. Tính khối lượng các chất thu được sau phản ứng.

2 câu trả lời

Đáp án + Giải thích các bước giải:

 Bài 1 :

a, Chất tác dụng với H2O là : CO2 , K2O , Al2O3

CO2 + H2O -> H2CO3

K2O + H2O -> 2KOH

Al2O3 + H2O -> Al(OH)3

b, Chất tác dụng với H2SO4 là : Fe , K2O , Al2O3, KOH

Fe + H2SO4 -> FeSO4 + H2

K2O + H2SO4 -> K2SO4 + H2O

Al2O3 + 3H2SO4 -> Al2(SO4)3 + 3H2O

2KOH + H2SO4 -> K2SO4 + 2H2O

c, Chất tác dụng với NaOH là : CO2 , Al2O3

CO2 + 2NaOH -> Na2CO3 + H2O

Al2O3 + 2NaOH -> 2NaAlO2 + H2O

Bài 2 :                                         Giải 

   nS = 3,2/32 = 0,1 (mol)

   nO2 = 4,48/22,4 = 0,2 (mol)

PTHH :    S   +   O2  -to->   SO2 

trc pứ :  0,1   -    0,2        -      0   (mol)

trg pứ :  0,1   -    0,1        -     0,1  (mol)

sau pứ :  0     -    0,1        -      0,1  (mol)

mO2 = 0,1 . 32 = 3,2 (g)

mSO2 = 0,1 . 64 = 6,4 (g)

chúc bn hc tốt cho mk 5* và trả lời hay nhất nhé 

Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
C{O_2} + {H_2}O \to {H_2}C{O_3}\\
{K_2}O + {H_2}O \to 2KOH\\
b)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{K_2}O + {H_2}S{O_4} \to {K_2}S{O_4} + {H_2}O\\
A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
c)\\
C{O_2} + 2NaOH \to N{a_2}C{O_3}\\
A{l_2}{O_3} + 2NaOH \to 2NaAl{O_2} + {H_2}O\\
2)\\
S + {O_2} \to S{O_2}\\
{n_S} = \dfrac{{3,2}}{{32}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = 0,2 - 0,1 = 0,1mol\\
{m_{{O_2}d}} = 0,1 \times 32 = 3,2g\\
{n_{S{O_2}}} = {n_S} = 0,1mol\\
{m_{S{O_2}}} = 0,1 \times 64 = 6,4g
\end{array}\)

 

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