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>
is directly related to the ideal gas law.
. Rewriting this gives:
. R is the universal gas constant, so its value is fixed. For a given sample, n is the number of moles of the gas, so its value would be fixed too. This means that the value of 
would be a fixed constant as well. Therefore, whatever the initial value 
is, it should be equal to the final value 
.</span>