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Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate - HSC - SSCE Chemistry - Question 19 - 2016 - Paper 1

Question 19

Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate. What is the mass of the solid formed? (A) 4.65 g (B) 8.69 g (C) 9.33 g (D) 31.5 g

Worked Solution & Example Answer:Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate - HSC - SSCE Chemistry - Question 19 - 2016 - Paper 1

Step 1

Determine the reaction between barium nitrate and sodium sulfate

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Answer

The reaction between barium nitrate (Ba(NO₃)₂) and sodium sulfate (Na₂SO₄) produces barium sulfate (BaSO₄) as a solid precipitate:

$Ba(NO_3)_2 + Na_2SO_4 \rightarrow BaSO_4 (s) + 2 NaNO_3$

Step 2

Calculate the moles of sodium sulfate

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Answer

The molarity (C) of sodium sulfate is given as 0.200 mol L⁻¹ and the volume (V) is 200 mL, which is equivalent to 0.200 L:

$n_{Na_2SO_4} = C \times V = 0.200 \, \text{mol L}^{-1} \times 0.200 \, \text{L} = 0.0400 \, \text{mol}$

Step 3

Determine the moles of barium sulfate formed

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Answer

Since the reaction between sodium sulfate and barium nitrate is a 1:1 ratio, the moles of barium sulfate formed will also be 0.0400 mol.

Step 4

Calculate the mass of barium sulfate formed

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Answer

The molar mass of barium sulfate (BaSO₄) is calculated as:

Molar mass of Ba = 137.33 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol (4 O atoms)

$\text{Molar mass of BaSO}_4 = 137.33 + 32.07 + (4 \times 16.00) = 233.39 \, \text{g/mol}$

Now, using the number of moles to find the mass:

$\text{mass} = n \times \text{molar mass} = 0.0400 \, \text{mol} \times 233.39 \, \text{g/mol} = 9.34 \, \text{g}$

Step 5

Conclusion

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Answer

Based on the calculations, the mass of the solid formed, which is barium sulfate, is approximately 9.34 g. Therefore, the correct answer is (C) 9.33 g.

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Questions answered

Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate - HSC - SSCE Chemistry - Question 19 - 2016 - Paper 1

Worked Solution & Example Answer:Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate - HSC - SSCE Chemistry - Question 19 - 2016 - Paper 1

Determine the reaction between barium nitrate and sodium sulfate

Calculate the moles of sodium sulfate

Determine the moles of barium sulfate formed

Calculate the mass of barium sulfate formed

Conclusion

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