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STatiana [176]
4 years ago
8

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
Chemistry
1 answer:
nata0808 [166]4 years ago
5 0

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) = <em>0.203moles NaHCO₃</em> ₓ (84g / 1mol NaHCO₃) =

<em>17.06g of NaHCO₃</em>

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

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

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 = <em>0.149mol CO₂</em>

<em />

These moles of CO₂ are produced from:

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

<em>15.84g of  </em>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) =

<em>9.02g of </em>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 =

<em>82.58g NaBr</em>

<em />

<em>Composition of initial mixture is:</em>

<em>9.02g of NaBrO₃</em>

<em>15.84g of  Na₂CO₃</em>

<em>17.06g of NaHCO₃</em>

<em>82.58g NaBr</em>

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1. The bar of metal is not pure gold.

2. The volume of 50.0 g of air is 42,016 mL.

3. The mass of the pure aluminum is 135 g.

4. The light beam's wavelength in meters is 475 x  10^{-9 m.

<h3>What is density?</h3>

The density of a substance is the ratio of the mass of the substance and its volume.

Density = mass/volume.

For the first problem: mass = 57.3 g, volume = 4.7 cm3

Density of the metal = 57.3/4.7

                                   = 12.19 g/cm3

The density of gold is 19.3 g/cm3. Thus, the bar of metal is not pure gold.

For the second problem: density = 1.19 x 10^{-3 g/ml, mass = 50.0 g

Volume = mass/density

                 = 50/1.19 x 10^{-3

                      = 42,016 ml

The volume of 50.0 g of air is 42,016 mL.

For the third problem: density = 2.70 g/ml, volume = 50.0 ml.

Mass = density x volume

                 = 2.70 x 50

                       = 135 g

The mass of the pure aluminum is 135 g.

Lastly:

nano = 10^{-9

475 nm = 475 x  10^{-9 m

The light beam's wavelength in meters is 475 x  10^{-9 m.

More on density can be found here: brainly.com/question/15164682

#SPJ1

6 0
1 year ago
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