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3 years ago

Calculate the density of a block of metal with a volume of 12.5 cm cubed and mass of 126.0 g

Chemistry

2 answers:

Step2247 [10]3 years ago

5 0

Answer : The density of the metal will be $10.08g/cm^3$

Explanation :

Density : It is defined as the mass of a substance contained per unit volume.

Formula used :

$Density=\frac{Mass}{Volume}$

Given:

Mass of metal = 126.0 g

Volume of metal = $12.5cm^3$

Now put all the given values in the above formula, we get:

$Density=\frac{126.0g}{12.5cm^3}=10.08g/cm^3$

Thus, the density of the metal will be $10.08g/cm^3$

SVEN [57.7K]3 years ago

3 0

Density is defined as mass/volume (or m/v).

So,

(126.0 g)/(12.5 cm^3)= 10.08 g/cm^3

If your teacher requires correct significant figures, the answer is 10.1 g/cm^3.
If not, the first answer is fine.

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Answer:

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Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

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From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

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4. Identify the limiting and excess reactants

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The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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