1/ 

a, $P_2O_5+3H_2O\to 2H_3PO_4$

b, 

$n_{P_2O_5}=\frac{14,2}{142}=0,1 mol$

$\Rightarrow n_{H_3PO_4}=0,2 mol$

$C_{M_{H_3PO_4}}=\frac{0,2}{0,2}=1M$

2/

a, $Na_2O+H_2O\to 2NaOH$ 

b, 

$n_{Na_2O}=\frac{6,2}{62}=0,1 mol$

$\Rightarrow n_{NaOH}=0,2 mol$

$m_{NaOH}=0,2.40=8g$

$C\%_{NaOH}=\frac{8.100}{120}= 6,67\%$

Đáp án:

1/

$b,CM_{H_3PO_4}=1M.$

2/

$b,C\%_{NaOH}=6,67\%$

Giải thích các bước giải:

1/

$a,PTPƯ:P_2O_5+3H_2O\xrightarrow{} 2H_3PO_4$

$b,n_{P_2O_5}=\dfrac{14,2}{142}=0,1mol.$

$Theo$ $pt:$ $n_{H_3PO_4}=2n_{P_2O_5}=0,2mol.$

$⇒CM_{H_3PO_4}=\dfrac{0,2}{0,2}=1M.$

2/

$a,PTPƯ:Na_2O+H_2O\xrightarrow{} 2NaOH$

$b,n_{Na_2O}=\dfrac{6,2}{62}=0,1mol.$

$Theo$ $pt:$ $n_{NaOH}=2n_{Na_2O}=0,2mol.$

$⇒m_{NaOH}=0,2.40=8g.$

$⇒C\%_{NaOH}=\dfrac{8}{120}.100\%=6,67\%$

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