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igomit [66]
3 years ago
5

I need help with this​

Chemistry
1 answer:
Darya [45]3 years ago
5 0

Answer:

the answer is the last one , sorry if im wrong.

Explanation:

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Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 57.0 g CO2. T
Alex_Xolod [135]

Answer:

26.9 g

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Explanation:

The equation of the reaction is;

4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)

Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles

4 moles of KO2 yields 2 moles of K2CO3

0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3

Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles

2 moles of CO2 yields 2 moles of K2CO3

1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3

Hence the limiting reactant is KO2

Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g

Percent yield = actual yield/theoretical yield × 100

Percent yield = 21.8/26.9 × 100

Percent yield = 81%

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4 years ago
If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of      sand, what  percent by mass of
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