\(PTHH:H3PO4+3NaOH\rightarrow Na3PO4+3H2O\)
__________1___________3_________1_________3_________(mol)
__________0,5________0,3_________0,1_______0,3________(mol)
Ta có :
\(n_{H3PO4}=0,25.2=0,5\left(mol\right)\)
\(n_{NaOH}=1,5.0,2=0,3\left(mol\right)\)
Ta thấy :\(\frac{0,5}{1}>\frac{0,3}{3}\)
\(\rightarrow\) NaOH hết ; H3PO4 dư
\(\Rightarrow m_{Na3PO4}=0,1.164=16,4\left(g\right)\)