$a,n_{HCl}=\dfrac{6,72}{22,4}=0,3mol.$
$⇒m_{HCl}=0,3.36,5=10,95g.$
$⇒C\%_{HCl}=\dfrac{10,95}{400+10,95}.100\%=2,66\%$
$V_{H_2O}=400ml=0,4l.$
$⇒CM_{HCl}=\dfrac{0,3}{0,4}=0,75M.$
$b,n_{NH_3}=\dfrac{6,72}{22,4}=0,3mol.$
$⇒m_{NH_3}=0,3.17=5,1g.$
$m_{H_2O}=300g.$
$⇒C\%_{NH_3}=\dfrac{5,1}{5,1+300}.100\%=1,67\%$
$⇒CM_{NH_3}=\dfrac{0,3}{0,3}=1M.$
$c,PTPƯ:Na_2O+H_2O\xrightarrow{} 2NaOH$
$n_{Na_2O}=\dfrac{15}{62}=0,24mol.$
$Theo$ $pt:$ $n_{NaOH}=2n_{Na_2O}=0,48mol.$
$⇒m_{dd\ spư}=15+400=415g.$
$⇒C\%_{NaOH}=\dfrac{0,48.40}{415}.100\%=4,6\%$
$V_{H_2O}=400ml=0,4l.$
$⇒CM_{NaOH}=\dfrac{0,48}{0,4}=1,2M.$
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