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Ann [662]
3 years ago
9

A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o

f H₂O, and some NO₂ . A possible empirical formula of the compound is ________.
Chemistry
1 answer:
koban [17]3 years ago
3 0

Answer:

CH₅N

Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

(45.0 g)(mol/18.02g) = 2.497...mol H₂O

Moles of H is found using the molar ratio of 2H:1H₂O:

(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

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Numbers of red line are on the right, numbers of blue line are on the left.

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A sample of carbon -12 has a mass of 6.00 g. How many atoms of carbon -12 are in the sample ?
nevsk [136]

Answer:

3.02e23

Explanation:

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4 0
3 years ago
Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:
kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g =  0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3  = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂   I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0
3 years ago
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