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Parents Pricing Home SSCE HSC Chemistry Chemical Reactions and Stoichiometry Excess barium nitrate solution is added to 200 mL of 0.200 mol L⁻¹ sodium sulfate
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
View full question 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 ... show full transcript
View marking scheme 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
Calculate the number of moles of sodium sulfate Only available for registered users.
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First, we can calculate the number of moles of sodium sulfate ( ext{Na}_2 ext{SO}_4) present in the solution using the formula:
e x t m o l e s = e x t c o n c e n t r a t i o n i m e s e x t v o l u m e ext{moles} = ext{concentration} imes ext{volume} e x t m o l es = e x t co n ce n t r a t i o n im ese x t v o l u m e
Given:
Concentration = 0.200 mol L⁻¹
Volume = 200 mL = 0.200 L
Thus,
e x t m o l e s o f N a 2 e x t S O 4 = 0.200 e x t m o l L − 1 i m e s 0.200 e x t L = 0.0400 e x t m o l ext{moles of Na}_2 ext{SO}_4 = 0.200 ext{ mol L}^{-1} imes 0.200 ext{ L} = 0.0400 ext{ mol} e x t m o l eso f N a 2 e x t SO 4 = 0.200 e x t m o l L − 1 im es 0.200 e x t L = 0.0400 e x t m o l
Identify the reaction and stoichiometry Only available for registered users.
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The reaction between barium nitrate ( ext{Ba(NO}_3 ext{)}_2) and sodium sulfate forms barium sulfate ( ext{BaSO}_4) as a precipitate:
e x t B a ( N O 3 e x t ) 2 + e x t N a 2 e x t S O 4 → e x t B a S O 4 ↓ + 2 e x t N a N O 3 ext{Ba(NO}_3 ext{)}_2 + ext{Na}_2 ext{SO}_4 \rightarrow ext{BaSO}_4 \downarrow + 2 ext{NaNO}_3 e x t B a ( NO 3 e x t ) 2 + e x t N a 2 e x t SO 4 → e x t B a SO 4 ↓ + 2 e x t N a NO 3
From the balanced equation, 1 mole of sodium sulfate reacts with 1 mole of barium nitrate to produce 1 mole of barium sulfate.
Calculate the moles of barium sulfate formed Only available for registered users.
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Since we have 0.0400 moles of sodium sulfate and it reacts in a 1:1 ratio with barium nitrate, we will also form 0.0400 moles of barium sulfate.
Calculate the mass of barium sulfate Only available for registered users.
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Now, we can calculate the mass of barium sulfate produced. The molar mass of barium sulfate ( ext{BaSO}_4) can be calculated as follows:
Atomic mass of Ba = 137.33 g/mol
Atomic mass of S = 32.07 g/mol
Atomic mass of O = 16.00 g/mol (×4)
So,
e x t M o l a r m a s s o f B a S O 4 = 137.33 + 32.07 + ( 16.00 i m e s 4 ) = 233.39 e x t g / m o l ext{Molar mass of BaSO}_4 = 137.33 + 32.07 + (16.00 imes 4) = 233.39 ext{ g/mol} e x t M o l a r ma sso f B a SO 4 = 137.33 + 32.07 + ( 16.00 im es 4 ) = 233.39 e x t g / m o l
Using the moles of barium sulfate:
e x t M a s s = e x t m o l e s i m e s e x t m o l a r m a s s = 0.0400 e x t m o l i m e s 233.39 e x t g / m o l = 9.33 e x t g ext{Mass} = ext{moles} imes ext{molar mass} = 0.0400 ext{ mol} imes 233.39 ext{ g/mol} = 9.33 ext{ g} e x t M a ss = e x t m o l es im ese x t m o l a r ma ss = 0.0400 e x t m o l im es 233.39 e x t g / m o l = 9.33 e x t g
Final Answer Only available for registered users.
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Thus, the mass of the solid formed, barium sulfate, is 9.33 g. Therefore, the correct answer is (C) 9.33 g.
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