3NaOH +H3PO4--->Na3PO4 + 3H2O
TH1 NaOH dư
Theo pthh
n\(_{Na3PO4}=n_{H3PO4}=b\left(mol\right)\)
m\(_{Na3PO4}=164b\left(g\right)\)
TH2: H3PO4 dư
Theo pthh
n\(_{Na3PO4}=\frac{1}{3}n_{NaOH}=0,33a\left(mol\right)\)
m\(_{Na3PO4}=0,33b.164=54,67\left(g\right)\)
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