If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) phosphate are produced?

Respuesta :

Answer:

If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) phosphate are produced?

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Explanation:

Answer:

164.2726 g

The ratio in the equation is 3:1 so the limit reaction ratio is 3:1 therefore,  0.392456 mol of were reaction.

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