Answer:
1.52g NaHCO3 were in the original mixture.
Mass percent: 64.1%
Explanation:
When NaHCO3 heats it descomposition occurs as follows:
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g).
The loss in mass is because of the evaporization of CO2 and H2O. As both are in the same porportion, its molar mass is the sum of both compounds (44g/mol + 18g/mol = 62g/mol)
Loss in mass: 74.80g - 74.24g = 0.56g.
In moles:
0.56g * (1mol / 62g) = 0.00903 moles of gas.
As 1 mole of the gases comes from 2 moles of NaHCO3:
<em>Moles NaHCO3:</em>
0.00903 moles of gas * (2 moles NaHCO3 / 1 mole gas) = 0.018 moles NaHCO3.
In grams (Molar mass NaHCO3: 84g/mol):
0.018 moles NaHCO3 * (84g / mol) = 1.52g NaHCO3 were in the original mixture.
The mass of the mixture was:
74.80g - 72.428g = 2.372g
That means mass percent of NaHCO3 is:
(1.52g / 2.372g) * 100 = 64.1%
