If you added an excess of sodium phosphate dodecahydrate to 2.122 g of barium chloride dihydrate, how many grams of barium phosphate could you synthesize? Note: Report your answer to three decimal places
The Correct Answer and Explanation is:
To solve this problem, we need to determine how many grams of barium phosphate (Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2) can be synthesized when an excess of sodium phosphate dodecahydrate (Na3PO4⋅12H2O\text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O}Na3PO4⋅12H2O) is added to 2.122 grams of barium chloride dihydrate (BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O}BaCl2⋅2H2O).
Step 1: Write the balanced chemical equation
The reaction between barium chloride and sodium phosphate can be represented as:3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl3 \text{BaCl}_2 + 2 \text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3\text{(PO}_4)_2 + 6 \text{NaCl}3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl
Step 2: Calculate the molar mass of BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O}BaCl2⋅2H2O
First, we calculate the molar mass of barium chloride dihydrate:
- Barium (Ba): 137.33 g/mol
- Chlorine (Cl): 35.45 g/mol (x2 for BaCl2\text{BaCl}_2BaCl2)
- Water (H₂O): 18.015 g/mol (x2 for 2H2O2\text{H}_2\text{O}2H2O)
So, the molar mass of BaCl2⋅2H2O\text{BaCl}_2 \cdot 2\text{H}_2\text{O}BaCl2⋅2H2O is:137.33+2(35.45)+2(18.015)=137.33+70.90+36.03=244.26 g/mol137.33 + 2(35.45) + 2(18.015) = 137.33 + 70.90 + 36.03 = 244.26 \, \text{g/mol}137.33+2(35.45)+2(18.015)=137.33+70.90+36.03=244.26g/mol
Step 3: Convert mass of barium chloride to moles
Next, we convert the mass of barium chloride dihydrate (2.122 g) to moles:Moles of BaCl2⋅2H2O=2.122 g244.26 g/mol=0.00868 mol\text{Moles of } \text{BaCl}_2 \cdot 2\text{H}_2\text{O} = \frac{2.122 \, \text{g}}{244.26 \, \text{g/mol}} = 0.00868 \, \text{mol}Moles of BaCl2⋅2H2O=244.26g/mol2.122g=0.00868mol
Step 4: Use the stoichiometry of the balanced equation
From the balanced equation, we know that 3 moles of BaCl2\text{BaCl}_2BaCl2 react with 2 moles of Na3PO4\text{Na}_3\text{PO}_4Na3PO4 to produce 1 mole of Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2.
The molar ratio of BaCl2\text{BaCl}_2BaCl2 to Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2 is 3:1. Therefore, the moles of Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2 produced is:Moles of Ba3(PO4)2=0.00868 mol BaCl23=0.00289 mol Ba3(PO4)2\text{Moles of } \text{Ba}_3\text{(PO}_4)_2 = \frac{0.00868 \, \text{mol} \, \text{BaCl}_2}{3} = 0.00289 \, \text{mol} \, \text{Ba}_3\text{(PO}_4)_2Moles of Ba3(PO4)2=30.00868molBaCl2=0.00289molBa3(PO4)2
Step 5: Calculate the mass of Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2
Next, we calculate the molar mass of barium phosphate:
- Barium (Ba): 137.33 g/mol (x3 for Ba3\text{Ba}_3Ba3)
- Phosphorus (P): 30.974 g/mol (x2 for PO4\text{PO}_4PO4)
- Oxygen (O): 16.00 g/mol (x8 for PO4\text{PO}_4PO4)
The molar mass of Ba3(PO4)2\text{Ba}_3\text{(PO}_4)_2Ba3(PO4)2 is:3(137.33)+2(30.974)+8(16.00)=411.99+61.948+128.00=601.938 g/mol3(137.33) + 2(30.974) + 8(16.00) = 411.99 + 61.948 + 128.00 = 601.938 \, \text{g/mol}3(137.33)+2(30.974)+8(16.00)=411.99+61.948+128.00=601.938g/mol
Now, we calculate the mass of barium phosphate produced:Mass of Ba3(PO4)2=0.00289 mol×601.938 g/mol=1.738 g\text{Mass of } \text{Ba}_3\text{(PO}_4)_2 = 0.00289 \, \text{mol} \times 601.938 \, \text{g/mol} = 1.738 \, \text{g}Mass of Ba3(PO4)2=0.00289mol×601.938g/mol=1.738g
Final Answer:
The amount of barium phosphate that could be synthesized is 1.738 g.
