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boyakko [2]
3 years ago
8

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C

3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O 2 ( g ) O2(g) are needed to completely burn 19.7 g C 3 H 8 ( g ) ?
Chemistry
1 answer:
Kipish [7]3 years ago
7 0

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

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baherus [9]

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8. What is the heat of reaction when hydrogen and oxygen combine to form water?
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<u>C. 571.6 kJ</u>

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Enthalpy Change = The enthalpy change for the formation of 1 mole of the substance from their standard state is called the enthalpy of formation.

This is intensive quantity as it is fixed for 1 mole .

Intensive properties = Those properties which are independent on the amount of the substance are intensive properties.

The value of these quantities does not get halve if you divide the substance into two equal parts. example , density, refractive index.

However , the enthalpy of reaction is extensive. Because on increasing the amount the value of the enthalpy also get doubles

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