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Math  /  Algebra

QuestionEXAMPLE: Determine the total number of ions that will be present in the precipitate formed when a solution with 20.00 grams of barium nitrate reacts with a solution containing excess potassium phosphate.

Studdy Solution

STEP 1

1. Barium nitrate reacts with potassium phosphate to form a precipitate.
2. The precipitate formed is barium phosphate.
3. The reaction goes to completion.
4. The molar masses of barium nitrate and barium phosphate are known.
5. We need to determine the total number of ions in the precipitate.

STEP 2

1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of barium nitrate.
3. Determine the moles of barium phosphate formed.
4. Calculate the total number of ions in the precipitate.

STEP 3

Write the balanced chemical equation for the reaction.
The balanced equation for the reaction between barium nitrate (Ba(NO3)2)(\text{Ba(NO}_3\text{)}_2) and potassium phosphate (K3PO4)(\text{K}_3\text{PO}_4) is:
3Ba(NO3)2+2K3PO4→Ba3(PO4)2↓+6KNO3 3\text{Ba(NO}_3\text{)}_2 + 2\text{K}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 \downarrow + 6\text{KNO}_3

STEP 4

Calculate the moles of barium nitrate.
The molar mass of barium nitrate (Ba(NO3)2)(\text{Ba(NO}_3\text{)}_2) is approximately 261.34 g/mol.
Calculate the moles of barium nitrate:
Moles of Ba(NO3)2=20.00 g261.34 g/mol≈0.0765 mol \text{Moles of Ba(NO}_3\text{)}_2 = \frac{20.00 \text{ g}}{261.34 \text{ g/mol}} \approx 0.0765 \text{ mol}

STEP 5

Determine the moles of barium phosphate formed.
From the balanced equation, 3 moles of barium nitrate produce 1 mole of barium phosphate (Ba3(PO4)2)(\text{Ba}_3(\text{PO}_4)_2).
Calculate the moles of barium phosphate:
Moles of Ba3(PO4)2=0.0765 mol3≈0.0255 mol \text{Moles of Ba}_3(\text{PO}_4)_2 = \frac{0.0765 \text{ mol}}{3} \approx 0.0255 \text{ mol}

STEP 6

Calculate the total number of ions in the precipitate.
Barium phosphate (Ba3(PO4)2)(\text{Ba}_3(\text{PO}_4)_2) dissociates into 3 barium ions (Ba2+)(\text{Ba}^{2+}) and 2 phosphate ions (PO43−)(\text{PO}_4^{3-}).
Total ions per formula unit of barium phosphate:
3+2=5 ions 3 + 2 = 5 \text{ ions}
Calculate the total number of ions:
Total ions=0.0255 mol×5 ions/mol×6.022×1023 ions/mol \text{Total ions} = 0.0255 \text{ mol} \times 5 \text{ ions/mol} \times 6.022 \times 10^{23} \text{ ions/mol}
Total ions≈7.68×1022 ions \text{Total ions} \approx 7.68 \times 10^{22} \text{ ions}

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