Chemistry, 19.01.2021 21:20 ewash04

C3H8 (g) 5O2 (g) -> 3CO2 (g) 4H2O (l) in the reaction represented baove, what is the total number of moles of reactants consumed when 1.00 mole of CO2 (g) is produced

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Answer from: loravillanueva87

The total number of moles of reactants consumed when 1.00 mole of CO₂ (g) is produced is 2 moles.

Explanation:

You have that the balanced reaction is:

C₃H₈ (g) + 5 O₂ (g) ⇒ 3 CO₂ (g) + 4 H₂O (l)

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

C₃H₈: 1 moleO₂: 5 molesCO₂: 3 molesH₂O: 4 moles

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So $x=\frac{c*b}{a}$

where a, b and c are known values. In this case the rule of three is applied as follows: if by stoichiometry of the reaction 3 moles of CO₂ consume 1 mole of C₃H₈, then 1 mole of CO₂ how many moles of C₃H₈ will it consume?

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

moles of C₃H₈= 0.33

Applying again a new rule of three: if by stoichiometry of the reaction 3 moles of CO₂ consume 5 moles of O₂, then 1 mole of CO₂ how many moles of O₂ will it consume?

$moles of O_{2} =\frac{1 mole of CO_{2} *5 moles of O_{2} }{ 3moles of CO_{2}}$

moles of O₂=1.67

The total number of moles of reactants consumed will be the sum of the consumed moles of C₃H₈ and the consumed moles of O₂:

0.33 moles + 1.67 moles=2 moles

The total number of moles of reactants consumed when 1.00 mole of CO₂ (g) is produced is 2 moles.

Answer from: Quest

i think it's letter a.

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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C3H8 (g) 5O2 (g) -> 3CO2 (g) 4H2O (l) in the reaction represented baove, what is the total number...

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