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2 years ago

24. A laboratory chemist wants to produce 25.0 g NO2 by the following reactions. If the

Chemistry

1 answer:

lyudmila [28]2 years ago

5 0

The amount of N2O5 to start with would be 35.79 grams

<h3>Stoichiometric calculations</h3>

From the balanced equation of the reactions:

$2N_2O_5 --- > 4NO_2 + O_2$

Mole ratio of N2O5 and NO2 = 1:2

Since the reaction's actual yield is 82%, the theoretical yield would be: 25 x 100/82 = 30.49 grams

Mole of 25 g NO2 = 30.49/46

= 0.66 mole

Equivalent mole of N2O5 = 0.54/2 mole

= 0.33 moles

Mass of 0.27 mole N2O5 = 0.33 x 108.01

= 35.79 grams

More on stoichiometric calculations can be found here: brainly.com/question/8062886

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3 years ago

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Read 2 more answers

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3 years ago

Determine the number of atoms in 51.0 grams of sodium, Na. (The mass of one mole of sodium is 22.99 g.)

miskamm [114]

Answer:

The answer is

<h3>1.340 × 10²⁴ sodium atoms</h3>

To find the number of atoms of sodium we use the formula

N = n × L

where

n is the number of moles

N is the number of entities

L is Avogadro's constant which is

$6.02 \times {10}^{23}$

We need to find the number of moles first

The formula is

$n = \frac{m}{M}$

where

M is the molar mass

m is the mass

n is the number of moles

From the question

M = 22.9 g/mol

m = 51.0 g

$n = \frac{51}{22.9}$

n = 2.227 moles

So the number of sodium atoms is

$N = 2.227 \times 6.02 \times {10}^{23}$

We have the final answer as

<h3>1.340 × 10²⁴ sodium atoms</h3>

Hope this helps you

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