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zavuch27 [327]
3 years ago
15

2C 2 H 6 +7O 2 ***>4CO 2 +6H 2 O if 7.0 g of C 2 H 6 react with 18 g of O 2 , how many grams of water will be produced

Chemistry
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

grams H₂O produced = 8.7 grams

Explanation:

Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)

               7g           18g                             ?g

Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water

Moles Reactants

moles C₂H₆ = 7g/30g/mol = 0.233mol

moles O₂ = 18g/32g/mol = 0.563mol

Limiting Reactant => (Test for Limiting Reactant)  Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.

moles C₂H₆/2 = 0.233/2 = 0.12

moles O₂/7 = 0.08

<u><em>Limiting Reactant is O₂</em></u>

Moles and Grams of H₂O:

Use Limiting Reactant moles (not division value) to calculate moles of H₂O.

moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield

grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O

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7 0
3 years ago
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Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
NARA [144]

Answer:

ΔH = q_{p}

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

q_{com} = q_{p}

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q_{com}  = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

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ΔQ = the internal heat absorbed by the system which in this case is q_{p}

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substituting into the heat change equation

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5 0
3 years ago
Consider the following data showing the initial rate of a reaction (A→products) at several different concentrations of A.
OlgaM077 [116]

Answer:

(1) order = 2

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Explanation:

Given the reaction:

A--------->Product

The rate constant relation for the reaction is given as:

R(i) = K [A]............(*)

Where R(I) is rate constant at different concentration of A.

Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results

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0.177 = K [0.60]ⁿ .........(3)

Dividing (2) by (1) and (3) by (1)

Gives:

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4 = 2ⁿ; 2² = 2ⁿ; n = 2

Similarly

0.177/0.011 = [0.60/0.15]ⁿ

16.09 = 4ⁿ

16.09 = 16 (approximately)

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