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stiks02 [169]
2 years ago
12

Solid carbon is known to react with oxygen gas to produce carbon dioxide. If a mass of 1.2g of carbon is burned in oxygen, 4.4g

of carbon dioxide is found to form. How many grams of oxygen gas reacted with carbon?
Chemistry
1 answer:
Elden [556K]2 years ago
7 0

Now ,

C + O2  → CO2

According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.

No of moles = mass of the substance/molecular mass of the substance.

In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.

No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.

No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles

Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.

Also number of moles of O2 = mass of O2÷ molar mass of O2

Substituting number of moles of O2 as 0.1 we get

mass of O2(x)  = Number of moles of O2 × Molar mass of O2

Mass of O2 (x) = 0.1 × 32= 3.2 g

Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.



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Answer ; The question is missing in some details, but here are he details ;

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Explanation:

The detaile calculation is as shown in the attachment.

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3 years ago
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There is no question
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Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

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In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

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ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

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ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

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ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
3 years ago
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8 0
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