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lbvjy [14]
1 year ago
9

A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2

and once for H20.
b) If 0.647 mole of oxygen in used in the burning of propane, how many moles of each of H2O are produced? How many moles of C3H8 are consumed?

The balanced chemical reaction:
1 C3 + 5 O2 = 3CO2 + 4 H2O
Chemistry
1 answer:
4vir4ik [10]1 year ago
5 0

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

\frac{4}{1}\times 3.85 mol=15.4 mol of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

\frac{1}{5}\times 0.647 mol=0.1294 mol of propane.

0.1294 moles of propane are consumed.

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5 0
2 years ago
Calculate ΔH∘ in kilojoules for the reaction of acetylene (C2H2) (ΔH∘f=227.4kJ/mol) with O2 to yield carbon dioxide (CO2) (ΔH∘f=
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Answer: The value of \Delta H^o for the reaction is, -2512.4 kJ

Explanation:

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2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})]

We are given:

\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=227.4kJ

Putting values in above equation, we get:

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A compound composed of only c and f contains 17.39 c by mass. what is its empirical formula
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Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

 

C = 17.39 / 12 = 1.45

F = 82.61 / 19 = 4.35

 

Divide the two by the smaller number, so divide by 1.45

 

C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

 

So the empirical formula is:

CF3 

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