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2 years ago

What is the number if moles of nitrogen gas which react with 18 g of magnesium to form magnesium nitride compound?(mg=24,N=14)

Chemistry

1 answer:

makvit [3.9K]2 years ago

6 0

Answer:

It is true. a) 0.25 mol

Explanation:

Hello there?

To begin solving this problem, you have to write down the chemical equation and make sure it is well balanced.

The chemical equation is;

3Mg(s) + N2(g) => Mg3N2(s)

1 mole of Mg = 24g

We have 18g of Magnesium (Mg) reacting with Nitrogen gas (N2)

From our equation,

Mole ratio = 3 : 1, (Mg : N2)

1 mol Mg = 24g

x mol Mg = 18g

x mol Mg = (18/24) = 0.75 mol Mg

But mole ratio = 3 : 1 (Mg : N2)

This means that 3 => 0.75 mol Mg

What about ratio 1 of N2?

N2 = (0.75 mol ÷ 3)/1

= 0.25 mol N2

I hope this helps you to understand better. Have a nice studies. 

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6 0

3 years ago

Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

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Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

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By Stoichiometry of the reaction:

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Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

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Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

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

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Putting values in above equation, we get:

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