Đáp án:

$a,CM_{CH_3COOH}=4M.$

$b,V_{H_2}=1,12l.$

$c,V_{O_2}=0,56l.$

Giải thích các bước giải:

$a,PTPƯ:2CH_3COOH+Mg\xrightarrow{} (CH_3COO)_2Mg+H_2↑$

$n_{(CH_3COO)_2Mg}=\dfrac{7,1}{142}=0,05mol.$

$Theo$ $pt:$ $n_{CH_3COOH}=2n_{(CH_3COO)_2Mg}=0,1mol.$

Đổi 25 ml = 0,025 lít.

$⇒CM_{CH_3COOH}=\dfrac{0,1}{0,025}=4M.$

$b,Theo$ $pt:$ $n_{H_2}=n_{(CH_3COO)_2Mg}=0,05mol.$

$⇒V_{H_2}=0,05.22,4=1,12l.$

$c,PTPƯ:2H_2+O_2\xrightarrow{t^o} 2H_2O$

$Theo$ $pt:$ $n_{O_2}=\dfrac{1}{2}n_{H_2}=0,025mol.$

$⇒V_{O_2}=0,025.22,4=0,56l.$

chúc bạn học tốt!

$2CH_3COOH+Mg\to (CH_3COO)_2Mg+H_2↑$

$n_{(CH_3COO)_2Mg}=7,1/142=0,05mol$

$\text{a/Theo pt :}$

$n_{CH_3COOH}=2.n_{(CH_3COO)_2Mg}=2.0,05=0,1mol$

$⇒C_{M_{CH_3COOH}}=\dfrac{0,1}{0,025}=4M$

$\text{b/Theo pt :}$

$n_{H_2}=n_{(CH_3COO)_2Mg}=0,05mol⇒V_{H_2}=0,05.22m4=1,12l$

$c/2H_2+O_2\overset{t^o}\to 2H_2O$

$\text{Theo pt :}$

$n_{O_2}=1/2.n_{H_2}=1/2.0,05=0,025mol$

$⇒V_{O_2}=0,025.22,4=0,56l$