7/
a/
$nP_2O_5=\frac{1,42}{142}=0,01$
$P_2O_5+3H_2O \to 2H_3PO_4$
0,01 0,02
$100ml=0,1lit$
$nH_3PO_4=0,1.2=0,2$
$CMH_3PO_4 sau=\frac{0,02+0,2}{0,1}=2,2M$
b/
$H_3PO_4+3NaOH \to Na_3PO_4+3H_2O$
0,22 0,66 0,22
$mNaOH=0,66.40=26,8g$
$mddNaOH=\frac{26,8}{40\%}=67g$
c/
$nNa_3PO_4.12H_2O=0,22$
⇒$nH_2O=0,22.12=2,64$
$mH_2O=2,64.18=47,52g$
$\%mH_2O=\frac{47,52}{0,4.380}.100=31,26\%$