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Ainat [17]
3 years ago
14

A mixture of sodium bicarbonate and ammonium bicarbonate is 75.9% bicarbonate by mass. what is the mass percent of sodium bicarb

onate in the mixture?
Chemistry
1 answer:
Lorico [155]3 years ago
8 0
Answer: 24.1%,  under below assumptions.

Justification:

The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:

- sodium bicarbonate, and
- ammonium bicarbonate

.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.

Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.

With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.

In formulas:

100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%

=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate

=> % sodium bicarbonate = 100% - 75.9% = 24.1%

Answer: 24.1%
</span>
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Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g)  ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

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