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4 years ago

Gasoline, if 9.00 liters has a mass of 6120 g. Find the density in g/cm^3

Chemistry

1 answer:

Ilya [14]4 years ago

5 0

Answer:

The density of gasoline is 0.68 g/cm³.

Explanation:

Given data:

Volume of gasoline = 9.00 L

Mass of gasoline = 6120 g

Density of gasoline = ?

Solution:

first of all we will convert the unit, L to mL.

9.00 L × 1000 mL / 1 L = 9000 mL

Formula:

d = m/v

d = density

m = mass

v= volume

d = 6120 g/ 9000 mL

d = 0.68 g/cm³ ( mL =cm³ )

Thus, density of gasoline is 0.68 g/cm³ .

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Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

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Answer for Part (1) :

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Answer for Part (2) :

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

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3 years ago

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Answer:

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