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2 years ago

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

Chemistry

1 answer:

ahrayia [7]2 years ago

6 0

Taking into account the reaction stoichiometry and the definition of STP, 5.4 L of carbon dioxide is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

<h3>Moles of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of C₃H₈ form 3 moles of CO₂, 0.080 moles of C₃H₈ form how many moles of CO₂?

$amount of moles of CO_{2} =\frac{0.080 moles of C_{3} H_{8}x3 moles of CO_{2} }{1 mole of C_{3} H_{8} }$

<u><em>amount of moles of CO₂= 0.24 moles</em></u>

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Volume of CO₂ produced</h3>

You can apply the following rule of three: If by definition of STP 1 mole of CO₂ occupies 22.4 L, 0.24 moles of CO₂ how much volume does it occupy?

$volume of CO_{2} =\frac{0.24 molesx22.4 L}{1 mole}$

<u><em>volume of CO₂= 5.376 L ≅ 5.4 L</em></u>

Finally, 5.4 L of carbon dioxide is produced.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

STP conditions:

<u>brainly.com/question/26364483</u>

<u>brainly.com/question/8846039</u>

<u>brainly.com/question/1186356</u>

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jeka94

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

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since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

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3 years ago

How many moles of nitroge are there in 50.0 g of nitrogen?

chubhunter [2.5K]

1.785714286 moles

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