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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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Answer:

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

Composition of initial mixture is:

  • 9.02g of NaBrO₃
  • 15.84g of  Na₂CO₃
  • 17.06g of NaHCO₃
  • 82.58g NaBr

For the reactions:

  1. 2NaBrO₃(s) ⟶ 2NaBr(s) + 3O₂(g)
  2. 2NaHCO₃(s) ⟶ Na₂O(s) + H₂O(g) + 2CO₂(g)
  3. Na₂CO₃(s) ⟶ Na₂O(s)+CO₂(g)
  • H₂O(g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

[tex]1.83g H_2O * (1 mol H_2O / 18.02g) * (2 mol NaHCO_3 / 1 mol H_2O) \\\\= 0.203moles NaHCO_3 * (84g / 1mol NaHCO_3) \\\\=17.06\text{ g of }NaHCO_3[/tex]

  • CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

[tex]15.51\text{ g } CO_2 * (1mol / 44.01g) = 0.352\text{ moles of } CO_2[/tex]

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:

[tex]0.149mol CO_2 * (1 mol Na_2CO_3 / 1 mol CO_2) * (106g / 1mol Na_2CO_3) \\\\=15.84\text{ g of } Na_2CO_3[/tex]

  • O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

[tex]2.87g O_2 * (1 mol O_2 / 32g) * (2 mol NaBrO_3 / 3 mol O_2) * (150.9g / 1mol NaBrO_3) \\\\=9.02\text{ g of} NaBrO_3[/tex]

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

Find more information about "Mass percentage" here:

brainly.com/question/5493941

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