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If 0.380 mol of barium nitrate is allowed to react with an excess of phosphoric acid, how many moles of barium phosphate could be formed?

Ba (NO3)2+ H3PO4 → Ba3(PO4)2+ HNO3

Sagot :

If 0.380 mol of barium nitrate is allowed to react with an excess of phosphoric acid, 0.127 moles of barium phosphate could be formed.

How to calculate number of moles?

The number of moles of a compound can be calculated stoichiometrically as follows:

Based on this question, the following chemical equation is given:

Ba (NO3)2+ H3PO4 → Ba3(PO4)2+ HNO3

The balanced equation is as follows:

3Ba(NO3)2 + 2H3PO4 → Ba3(PO4)2 + 6HNO3

3 moles of barium nitrate produces 1 mole of barium phosphate

Therefore, 0.380 moles of barium nitrate will produce 0.380/3 = 0.127moles of barium phosphate.

Learn more about stoichiometry at: https://brainly.com/question/9743981

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