A mixture of NaBrO 3, NaBrO3, NaHCO 3, NaHCO3, Na 2 CO 3, Na2CO3, and NaBr NaBr was heated, producing H 2 O, H2O, CO 2, CO2, and O 2 O2 gases by the following equations. 2 NaBrO 3 (s) ⟶ 2 NaBr (s) + 3 O 2 (g) 2NaBrO3 (s) ⟶2NaBr (s) + 3O2 (g) 2 NaHCO 3 (s) ⟶ Na 2 O (s) + H 2 O (g) + 2 CO 2 (g) 2NaHCO3 (s) ⟶Na2O (s) + H2O (g) + 2CO2 (g) Na 2 CO 3 (s) ⟶ Na 2 O (s) + CO 2 (g) Na2CO3 (s) ⟶Na2O (s) + CO2 (g) If 124.5 g 124.5 g of the mixture produces 1.83 g 1.83 g of H 2 O, H2O, 15.51 g 15.51 g of CO 2, CO2, and 2.87 g 2.87 g of O 2, O2, what was the mass of each compound in the mixture? Assume complete decomposition of the mixture. NaBr NaBr does not react under the reaction conditions. mass of NaBrO 3 : NaBrO3: g mass of NaHCO 3 : NaHCO3: g mass of Na 2 CO 3 : Na2CO3: g mass of NaBr : NaBr: g

Question

Guadalupe Lamb

Chemistry

28 October, 22:32

A mixture of NaBrO 3, NaBrO3, NaHCO 3, NaHCO3, Na 2 CO 3, Na2CO3, and NaBr NaBr was heated, producing H 2 O, H2O, CO 2, CO2, and O 2 O2 gases by the following equations. 2 NaBrO 3 (s) ⟶ 2 NaBr (s) + 3 O 2 (g) 2NaBrO3 (s) ⟶2NaBr (s) + 3O2 (g) 2 NaHCO 3 (s) ⟶ Na 2 O (s) + H 2 O (g) + 2 CO 2 (g) 2NaHCO3 (s) ⟶Na2O (s) + H2O (g) + 2CO2 (g) Na 2 CO 3 (s) ⟶ Na 2 O (s) + CO 2 (g) Na2CO3 (s) ⟶Na2O (s) + CO2 (g) If 124.5 g 124.5 g of the mixture produces 1.83 g 1.83 g of H 2 O, H2O, 15.51 g 15.51 g of CO 2, CO2, and 2.87 g 2.87 g of O 2, O2, what was the mass of each compound in the mixture? Assume complete decomposition of the mixture. NaBr NaBr does not react under the reaction conditions. mass of NaBrO 3 : NaBrO3: g mass of NaHCO 3 : NaHCO3: g mass of Na 2 CO 3 : Na2CO3: g mass of NaBr : NaBr: g

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Answers (1)

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Dwayne Olsen · Answer 1

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

Explanation:

For the reactions:

2NaBrO₃ (s) ⟶ 2NaBr (s) + 3O₂ (g)

2NaHCO₃ (s) ⟶ Na₂O (s) + H₂O (g) + 2CO₂ (g)

Na₂CO₃ (s) ⟶ Na₂O (s) + CO₂ (g)

All H₂O (g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

1.83g H₂O ₓ (1 mol H₂O / 18.02g) ₓ (2 mol NaHCO₃ / 1 mol H₂O) = 0.203moles NaHCO₃ ₓ (84g / 1mol NaHCO₃) =

17.06g of NaHCO₃

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

15.51g CO₂ ₓ (1mol / 44.01g) = 0.352moles of CO₂

As 2 moles of NaHCO₃ produce 2 moles of CO₂, moles of CO₂ that comes from NaHCO₃ are 0.203moles NaHCO₃. Moles of CO₂ that comes from Na₂CO₃ are:

0.352mol CO₂ - 0.203mol CO₂ = 0.149mol CO₂

These moles of CO₂ are produced from:

0.149mol CO₂ ₓ (1 mol Na₂CO₃ / 1 mol CO₂) ₓ (106g / 1mol Na₂CO₃) =

15.84g of Na₂CO₃

And all O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

2.87g O₂ ₓ (1 mol O₂ / 32g) ₓ (2 mol NaBrO₃ / 3 mol O₂) ₓ (150.9g / 1mol NaBrO₃) =

9.02g of NaBrO₃

If initial mass of the mixture was 124.5g, mass of NaBr was:

124.5g - 9.02g of NaBrO₃ - 15.84g of Na₂CO₃ - 17.06g of NaHCO₃ =

82.58g NaBr

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr