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Radda [10]
2 years ago
10

The elemental composition of a compound is determined to be 39.5% oxygen, 32.1% chromium and 28.4% sodium by weight.

Chemistry
1 answer:
maksim [4K]2 years ago
3 0

Answer:

Explanation:

To calculate percent composition, we divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

​%C=

​12.04g compound

​

​7.34g C

​​ ×100%=61.0%

​%H=

​12.04g compound

​

​1.85g H

​​ ×100%=15.4%

​%N=

​12.04g compound

​

​2.85g N

​​ ×100%=23.7%

​​  

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

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The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains
andriy [413]

Answer: 67.2 moles

Explanation: 2C_2H5+5O_2\rightarrow 4CO_2+2H_2O

According to the given balanced equation, 2 moles of acetylene C_2H_2 combine with 5 moles of oxygen O_2  to produce 4 moles of carbon dioxide CO_2.

Thus if 2 moles of acetylene C_2H_2 combine with = 5 moles of oxygen O_2

35 moles of acetylene C_2H_2 combine with=\frac{5}{2}\times {35}=87.5 moles of oxygen O_2

But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen O_2 reacts with = 2 moles of acetylene C_2H_2

84 moles of oxygen O_2 reacts with=\frac{2}{5}\times {84}=33.6 moles of acetylene C_2H_2

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene C_2H_2  produce= 4 moles of carbon dioxide CO_2.

33.6 moles of acetylene C_2H_2 produce=\frac{4}{2}\times {33.6}=67.2 moles of of carbon dioxide CO_2.


8 0
3 years ago
Read 2 more answers
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