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30. (10 points) Magnesium nitride forms when magnesium reacts with nitrogen gas.

ID: 718224 • Letter: 3

Question

30. (10 points) Magnesium nitride forms when magnesium reacts with nitrogen gas. If 18.4 go magnesium nitride forms from the reaction of 20.0 g of magnesium and 10.0 g of nitrogen, what is percentage yield? the Balanced Chemical Equation: Limiting reactant: Theoretical Yield: Excess Reactant: Actual Yield: Percentage Yield: Explain 2 possibilities why this reaction does not have a 100% yield. . Assuming 100% yield, how many grams of the excess reactant will remain? 31. (5 points) 255 mL of vinegar (acetic acid HC2H302) spills and it takes 25.0 grams of baking soda (NaHCO3) to neutralize it. What is the concentration of the vinegar? NaHCO3+HC2H302 H20+CO2+NaC2H302

Explanation / Answer

Ans 30

3Mg + N2 = Mg3N2

Magnesium + nitrogen = magnesium nitride

Moles of Mg3N2 = mass/molecular weight

= (18.4g) / (100.93 g/mol)

= 0.1823 mol

Moles of Mg = mass/molecular weight

= (20g / 24.305g/mol)

= 0.8229 mol

Moles of N2 = mass/molecular weight

= (10g / 28.01g/mol)

= 0.3570 mol

From the stoichiometry of the reaction

Moles of N2 required = 1 mol N2 x 0.8229 mol Mg / 3 mol Mg

= 0.2743 mol

We have more moles of N2 (0.3570 mol) than required ( 0.2743 mol)

Excess reactant = N2

Limiting reactant = Mg

Moles of Mg3N2 produced

= 1 mol Mg3N2 x 0.8229 mol Mg / 3 mol Mg

= 0.2743 mol Mg3N2

Theoretical yield = moles x molecular weight

= 0.2743 mol x 100.92 g/mol

= 27.68 g

% yield = actual yield x 100 / theoretical yield

= 18.4 x 100 / 27.68

= 66.47 %

Two possibilities are

There is always loss of product while the reaction takes place due to which less product is obtained at the end.

In the process, some product is wasted into the filter paper or solid left behind, while transferring the solution from one beaker to another.

Human error is also a main factor of less yield. Some measurements can be done in less accuracy and less precise manner.

If 100% yield occured

Moles of Mg3N2 produced = 0.2743 mol

Excess reactant remain = Initial - reacted

= 0.3570 - 0.2743

= 0.0827 mol x 28.01g/mol

= 2.32 g

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