Consider the reaction
H3PO4+ 3NaOH→Na3PO4+ 3H2O.
How much Na3PO4 can be prepared by the reaction of 3.43 of H3PO4 with an excess of NaOH?
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a compound has a molecular mass of 99 and contains 24.8percent carbon, 3.9percent hydrogen and 71.3percent chlorine. Determine the empirical formula and the molecula formula
To determine how much Na3PO4 can be prepared from the given reaction, we need to use the concept of stoichiometry. The balanced chemical equation gives us the ratio of moles between the reactants and products.
The balanced equation is: H3PO4 + 3NaOH → Na3PO4 + 3H2O
From the balanced equation, we can see that for every 1 mole of H3PO4, we produce 1 mole of Na3PO4. This means the mole ratio of H3PO4 to Na3PO4 is 1:1.
Now, let's calculate the number of moles of H3PO4 available in the reaction:
Given mass of H3PO4 = 3.43 g
Molar mass of H3PO4 = 98.0 g/mol
Number of moles of H3PO4 = mass / molar mass
= 3.43 g / 98.0 g/mol
= 0.035 moles
Since the mole ratio is 1:1 between H3PO4 and Na3PO4, we can conclude that 0.035 moles of Na3PO4 can be prepared from 0.035 moles of H3PO4.
Therefore, the amount of Na3PO4 that can be prepared by the reaction of 3.43 g of H3PO4 with an excess of NaOH is 0.035 moles.