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Luda [366]
3 years ago
12

When 0.105 mol propane, C3H8 is burned in an excess of oxygen, how many moles of oxygen are consumed? The products are carbon di

oxide and water. You have to properly balance the equation before you can answer this question.
A) 0.420 mol O2
B) 0.525 mol O2
C) 0.875 mol O2
D) 0.905 mol O2
E) 0.975 mol O2
Chemistry
1 answer:
torisob [31]3 years ago
6 0

<u>Answer:</u> The amount of oxygen gas consumed is 0.525 moles

<u>Explanation:</u>

We are given:

Moles of propane burned = 0.150 moles

The chemical equation for the combustion of propane follows:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen gas

So, 0.150 moles of propane will react with = \frac{5}{1}\times 0.105=0.525mol of oxygen gas

Hence, the amount of oxygen gas consumed is 0.525 moles

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Answer:

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Explanation:

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<u>D to E</u>

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3 0
2 years ago
Most geologists think that the movement of Earth's plates is caused by
butalik [34]

Answer: C. Convection currents in the asthenosphere.

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7 0
3 years ago
If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions
dimulka [17.4K]

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

P_2 = final pressure of dry gas at STP =  760 mm Hg

V_1 = initial volume of dry gas = 85.0 mL

V_2 = final volume of dry gas at STP = ?

T_1 = initial temperature of dry gas = 20^oC=273+20=293K

T_2 = final temperature of dry gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get the final volume of wet gas at STP

\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}

V_2=77.4mL

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5 0
3 years ago
7th grade Sem 1
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4 0
3 years ago
Read 2 more answers
1. A mixture contains 8.00 g each of O2, CO2, and SO2 at STP. Calculate the volume of this mixture. Which of the gases would exe
Gekata [30.6K]

Answer:

Explanation:

mole of O₂ = \frac{8}{32}

= .25 moles

mole of CO₂

= \frac{8}{44}

= .1818 moles

moles of SO₂

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= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)

= 12.47 liter.

gas will exert partial pressure according to their mole fraction

gas having greatest no of moles in the total mole will have greatest mole fraction so

O₂ will have greatest partial pressure.

7 0
3 years ago
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