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kramer
3 years ago
9

Given a 1.50 Liters of solution that contains 62.5 grams of magnesium oxide. What is the Molarity of that solution? Please show

all work for any credit.
Chemistry
1 answer:
zhenek [66]3 years ago
3 0
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol

Molarity: mol/L

First, find mol.

62.5 g x 1mole ÷ 40.31 g = 1.55 mol

then divide mol and the given liters

1.55mol ÷ 1.50 L= 1.03 M
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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
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Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

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Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

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

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

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