Question

How many grams of h3po4 is required to completely neutralize 120g of naoh

Answer 1

Answer:

Chemical Equation:

H3PO4 + 3NaOH --->Na3PO4 + 3H2O

Here molecular mass of H3PO4 is 98u

NaOH molecular mass is 40u

Na3PO4 MOLECULAR mass is 164u and water MOLECULAR mass is 18u

So we get,

H3PO4+3NaOH ---->Na3PO4 + 3H2O

98g 120g 164g 54g

Therefore 98g of H3PO4 is required to completely neutralise 120g of NaOH.

Answer 2

Answer:

98 mol

Explanation:

120gms of NaOH=120/40=3 mol

3NaOH+H3PO4⇒Na3PO4+3H2O

3 moles of   NaOH reacts with 1 mole of H3PO4

molar mass of H3PO4=98gm/mol

mass of H3PO4=1 mol x 98gm/mol

                          =98mol

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