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3 years ago

How many formula units are in 9.039 moles of sodium chloride

Chemistry

1 answer:

hodyreva [135]3 years ago

7 0

Answer:

5.443 formula units

Explanation:

use avagadro's number: 1 mole = 6.022 × $10^{23}$ particles (in this case formula units)

number of mole × 6.022 × $10^{23}$ = number of particles

9.039 moles × 6.022 × $10^{23}$ = 5.443 × $10^{24}$ formula units

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3 years ago

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g).

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Answer: 2.1 g mass of ozone( $O_{3}$ ) is predicted to form from the reaction of 2.0 g $NO_{2}$ in a car's exhaust and excess oxygen

Given information : Mass of $NO_{2}$ = 2.0 g and $O_{2}$ is in excess.

We need to calculate the mass of ozone ( $O_{3}$ )

Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

$NO_{2} + O_{2}\rightarrow NO + O_{3}$

Step 1 : Convert grams of $NO_{2}$ to moles of $NO_{2}$ .

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $NO_{2}$ = 46.0 g/mol

$Moles = \frac{2.0g}{46.0\frac{g}{mol}}$

Moles of $NO_{2}$ = 0.043 mol

Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

A mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

Moles of $O_{3}$ = $(0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}$

Moles of $O_{3}$ = $(0.043)\times \frac{(1 mol O_{3})}{(1)}$

Moles of $O_{3}$ = 0.043 mol

Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

Grams = $(0.043)\times (\frac{48 g O_{3}}{1})$

Grams = 2.1 g $O_{3}$

Note : The above three steps can also be done using a single step setup.

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Grams of $O_{3}$ = $(2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}$

Grams of $O_{3}$ = 2.1 grams

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3 years ago