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Pachacha [2.7K]
3 years ago
12

In the combustion of ethane, shown above, how many moles of carbon dioxide can be produced from 1.00 mole of ethane? A. 0.500 mo

le B. 1.00 mole C. 2.00 moles D. 4.00 moles
Chemistry
1 answer:
Daniel [21]3 years ago
6 0
The balanced chemical equation for the <span>combustion of ethane is 
2C</span>₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,
    moles of 
CO₂(g) produced = moles of reacted C₂H₆(g) x 2
                                                 
= 1.00 mol x 2
                                                  = 2.00 mol

Hence, the correct answer is "C".
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3 years ago
3.00 L of a gas is collected at 35.0°C and 705.0 mmHg. What is the volume at STP?
makvit [3.9K]

Answer:

2.47L

Explanation:

Using the combined gas law equation as follows:

P1V1/T1= P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 705mmHg

P2 = 760mmHg (STP)

V1 = 3.00L

V2 = ?

T1 = 35°C = 35 + 273 = 308K

T2 = 273K (STP)

Using P1V1/T1= P2V2/T2

705 × 3/308 = 760 × V2/273

2115/308 = 760V2/273

Cross multiply

308 × 760V2 = 2115 × 273

234,080V2 = 577,395

V2 = 577,395 ÷ 234,080

V2 = 2.47L

8 0
3 years ago
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