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lara [203]
3 years ago
8

In a certain experiment 10.0g of magnesium nitride is allowed to react with 5.00 g of water. Calculate the final mass of ammonia

, water, and magnesium nitride at the end of the reaction.
Chemistry
1 answer:
valkas [14]3 years ago
4 0

Answer:

The final mass of ammonia is 1.57 grams

The final mass of water is 0 grams

The final mass of magnesium nitride is 5.34 grams

Explanation:

Step 1: Data given

Mass of Mg3N2 = 10.0 grams

Molar mass of Mg3N2 = 100.95 g/mol

Mass of H2O = 5.00 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

Step 3: Calculate moles Mg3N2

Moles Mg3N2 = mass / molar mass

Moles Mg3N2 = 10.0 grams / 100.95 g/mol

Moles Mg3N2 = 0.0991 moles

Step 4: Calculate moles H20

Moles H2O = 5.00 grams / 18.02 g/mol

Moles H2O = 0.277 moles

Step 5: Calculate the limiting reactant

H2O is the limiting reactant. It will completely be consumed (0.277 moles).

Mg3N2 is in excess. There will react 0.277/6 = 0.0462 moles

There will remain 0.0991 - 0.0462 = 0.0529 moles

This is 0.0529 moles * 100.95 g/mol = 5.34 grams

Step 6: Calculate moles of NH3

For 1 mol Mg3N2 we need 6 moles H2O to produce 3 moles Mg(OH)2 and 2 moles NH3

For 0.277 moles H2O we'll have 0.277/3 = 0.0923 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.0923 moles * 17.03 g/mol

Mass NH3 = 1.57 grams

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Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

2SO_2 + O_2 \to 2SO_3

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

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For O₂; we have 2.6 - x

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Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

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Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

Then:

K_c = (0.8325)^{1/2}

\mathbf{K_c = 0.912}

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

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