A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silver nitrate,...
Question:
A sample of 70.5 mg of potassium phosphate is added to 15.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. Assuming the reaction goes to completion, calculate the amount, in grams, of precipitate that forms.
{eq}K_3PO_4(aq) + 3AgNO_3(aq) \to Ag_3PO_4(s) + 3KNO_3(aq) {/eq}
Mole to Mole Relation:
When according to the given balanced reaction and reactant amount, the product amount has to be determined, then one should use the mole comparison to determine the product mole. Then mole formula will be helpful to calculate the amount (gram) of the product.
Answer and Explanation: 1
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View this answerGiven Data:
- The mass of potassium phosphate is 70.5 mg.
- The volume of silver nitrate is 15.0 mL and molarity is 0.050 M.
The mg amount is...
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Mole-to-Mole Ratios and Calculations of a Chemical Equation
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Chapter 9 / Lesson 2
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Learn about the mole ratio. Understand the definition of mole ratio, how to find mole ratio in stoichiometry, and see examples of using mole ratio in problems.
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