A sample of 70.5 mg of potassium phosphate is added to 15.0mL of 15.0 mL of 0.05
ID: 625576 • Letter: A
Question
A sample of 70.5 mg of potassium phosphate is added to 15.0mL of 15.0 mL of 0.050 M silver nitrate, resulting the formation of precipitate. Assuming the reaction of a precipitate. Assuming the reaction goes to completion, calculate the amount, in grams, of precipitate that forms. K3PO4(aq)+3AgNO3(aq)---->Ag3PO4(s)+3KNO3(aq)Explanation / Answer
follow this Determine the formula for the salts provided based on the charge of the ions (I put charge in parentheses): Ag(+), NO3(-), K(+), PO4(3-) Therefore, the formulas are AgNO3 and K3PO4 2. Decide what the products will be based on the type of reaction and the charges of the ions. The reaction is a double replacement (or precipitation) reaction, so the positive ions switch partners. Remember to write correct formulas for the products based on the charges. AgNO3 + K3PO4 --> KNO3 + Ag3PO4 3. Balance the equation. 3 AgNO3 + K3PO4 --> 3 KNO3 + Ag3PO4 4. Indicate the state of each substance (I'll assume the original reactants were aqueous solutions). The solid product is silver phosphate (all salts with potassium are soluble in water). 3 AgNO3 (aq) + K3PO4 (aq) --> 3 KNO3 (aq) + Ag3PO4 (s) 5. The soluble salts split up into separate ions when dissolved in water. Write the ionic equation: 3 Ag(+) + 3 NO3(-) + 3 K(+) + PO4(3-) --> 3 K(+) + 3 NO3(-) + Ag3PO4 (s) 6. Eliminate the ions that are EXACTLY the same on both sides to get the net ionic equation: 3 Ag(+) + PO4(3-) --> Ag3PO4 (s)