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aleksklad [387]
3 years ago
5

If 42 grams of carbon and 52 grams of oxygen are used, how many grams of CO2 will be produced (Hint: find the

Chemistry
1 answer:
zavuch27 [327]3 years ago
8 0

Answer:

71.5g

Explanation:

The reaction equation is given as:

               C  +  O₂  →  CO₂

Mass of C = 42g

Mass of O₂  = 52g

Unknown:

Mass of CO₂ produced  = ?

Solution

Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.

 The extent of the reaction is controlled by this reactant.

Find the number of moles of the given species;

 Number of moles  = \frac{mass}{molar mass}

      Number of moles of C  = \frac{42}{12}   = 3.5mol

     Number of moles of O₂   = \frac{52}{32}   = 1.63mol

Now;

   From the balanced reaction equation;

           1 mole of C reacted with 1 mole of O₂

We see that C is in excess and O₂ is the limiting reactant.

            1 mole of O₂ will produce 1 mole of CO₂

 So;      1.63mole of O₂ will produce 1.63 mole of CO₂

Mass of CO₂ = number of moles x molar mass

       Molar mass of CO₂ = 44g/mol

Mass of CO₂ = 1.63 x 44 = 71.5g

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The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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3 years ago
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