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mixer [17]
3 years ago
13

Consider the reaction below. If you start with 2.00 moles of C3H8 (propane) and 2.00 moles of O2, how many moles of carbon dioxi

de can be produced? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g). A) 8.00. B) 1.20. C) 6.00. D) 2.00.
Chemistry
1 answer:
Vladimir79 [104]3 years ago
5 0
From the balanced chemical reaction, the ratio of moles of propane to moles CO2 produced is 1/3. Given 2 moles of propane, 6 moles of CO2 is produced. Further, the ratio of moles of O2 to moles of CO2 produced is 5/3. Given 2 moles of oxygen, 1.2 moles of CO2 is produced. We take the smaller amount of CO2 as the answer. Thus, the answer is letter B. 1.2 moles
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Is the genie in the bottle experiment a physical or chemical change/reaction?
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Answer:

Chemical reaction.

Step-by-step explanation:

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2H₂O₂(ℓ) ⟶ 2H₂O(ℓ) + O₂(g)

New substances are formed, and old ones disappear, so this is a chemical reaction.

The reaction also releases a <em>large amount of heat</em>.

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How many electrons does the element Zn have?
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Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
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