If 8.23 g of magnesium chloride react completely with sodium phosphate, how many grams of magnesium phosphateare produced

The correct answer is 7.57 grams of magnesium phosphate.

Explanation:

Based on the given question, the chemical reaction taking place is:  

2Na₃PO₄ (aq) + 3MgCl₂ (aq) ⇒ Mg₃(PO₄)2 (s) + 6NaCl (aq)

From the given reaction, it is evident that two moles of sodium phosphate reacts with three moles of magnesium chloride to produce one mole of magnesium phosphate.  

Based on the given information, 8.23 grams of magnesium chloride reacts completely with sodium phosphate, therefore, magnesium chloride in the given case is the limiting reagent.  

In the given case, 3 moles of magnesium chloride produce 1 mole of magnesium phosphate. Therefore, 1 mole of magnesium chloride will produce 1/3 mole of magnesium phosphate.  

The molecular mass of magnesium chloride is 95.21 grams per mole. So, 1 mole of magnesium chloride is equivalent to 95.21 grams of magnesium chloride.  

On the other hand, the molecular mass of magnesium phosphate is 262.85 grams per mole. Therefore, 1 mole of magnesium phosphate is equal to 262.85 grams of magnesium phosphate.  

As seen earlier that 1 mole of magnesium chloride = 1/3 moles of magnesium phosphate. So,  

95.21 grams of magnesium chloride = 1/3 × 262.85 grams of magnesium phosphate

= 262.85 / 3 grams of magnesium phosphate

1 gram of magnesium chloride = 262.85 / 3 × 95.21 grams of magnesium phosphate

8.23 grams of magnesium chloride = 262.85 / 3 × 95.21 × 8.23 grams of magnesium phosphate

= 7.57 grams of magnesium phosphate

Hence, when 8.23 grams of magnesium chloride when reacts completely with sodium phosphate, it produces 7.57 grams of magnesium phosphate.  

c is the answer because when something is left after something evaporates that shows the chemical change.

the answer for sodium is 0.186

the answer for rubidium is 0.251