Set up ALL WORK in your notebook. 1. Empirical formula problems: (Use class note

Question

Set up ALL WORK in your notebook. 1. Empirical formula problems: (Use class notes!!) a. If a compound contains 2.08g C and 2.77g O, calculate the empirical formula. b. If a Compound contains 6.79g C and 1.13g H, calculate the empirical formula. c. If a compound contains 0.256g P and 3.294g Br, calculate the empirical formula. 2. Balance the equation. Na + H2O right arrow NaOH + H2 a. How many moles of Na must be reacted to produce 10.0g of hydrogen gas? b. If 36.0g of water reacts, how many moles of sodium will be consumed? 3. Balance the equation. Al + Zn(NO3)2 right arrow Al(NO3)3 + Zn Reaction of 5.00kg Al will result in the formation of how many grams of Zn? 4. Balance the equation. C3H8 + O2 right arrow CO2 + H2O In order to produce 300.g carbon dioxide, what mass of propane must be reacted? 5. Balance the equation. H2 + O2 right arrow H20 a. If 45 mg of cadi reactant are combined, which reactant is the limiting reagent? b. What mass of water will be made? 6.Be able to define percent yield, limiting reagent, excess reagent. 7.If the predicted yield for a product in chemical reaction is 45.8g, and the actual amount of product collected equals 37.5g, calculate the percent yield. 8.Review your most recent lab calculations. 9.Take the section quizzes in the book on p. 789, 801.

Explanation / Answer

1.

a.

Mass of elements: This is given in the question.

C = 2.08 g

O = 2.77 g

Moles of elements:

We need to divide the masses given with their respective atomic weights.

C = 2.08 / 12 = 0.173 mol

O = 2.77 / 16 = 0.173 mol

Mole ratio of elements:

Here, we need to divide the moles found with the least number of moles.

C = 1.73 / 1.73 = 1

O = 1.73 / 1.73 = 1

So, the ratio C:O = 1:1

So, the empirical formula is CO

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b.

Mass of elements: This is given in the question.

C = 6.79 g

H = 1.13 g

Moles of elements:

We need to divide the masses given with their respective atomic weights.

C = 6.76 / 12 = 0.565 mol

H = 1.13 / 1 = 1.13 mol

Mole ratio of elements:

Here, we need to divide the moles found with the least number of moles.

C = 0.565 / 0.565 = 1

H = 1.13 / 0.565 = 2

So, the ratio C:H = 1:2

So, the empirical formula is CH2

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c.

Mass of elements: This is given in the question.

P = 0.256

Br = 3.294 g

Moles of elements:

We need to divide the masses given with their respective atomic weights.

P = 0.256 / 31 = 8.258 x 10^-3

Br = 3.294 / 80 = 0.0412

Mole ratio of elements:

Here, we need to divide the moles found with the least number of moles.

P = 8.258 x 10^-3 / 8.258 x 10^-3 = 1

Br = 0.0412 / 8.258 x 10^-3 = 5

So, the ratio P : Br = 1:5

So, the empirical formula is PBr5

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2.

Balancing the equation

2Na + 2H2O ----------------> 2NaOH + H2

a.

Mass of hydrogen gas = 10.0 g

Moles of hydrogen gas = 10.0 / 2 = 5.0 mol

from the balanced equation, it is clear that 1 mol of H2 is produced from 2 mol of Na.

So, 5.0 mol (10.0g) of H2 is produced from 10.0 mol of Na.

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b.

mass of water = 36.0 g

Moles of water = 36.0 / 2 = 18.0 mol

From the balanced equation, it is clear that, Na and H2O react in 2:2 that is equimolar ratio.

So, MOles of water = Moles of Na = 18.0 mol.

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3.

Balancing the equation

2Al + 3Zn(NO3)2 ---------> 2Al(NO3)3 + 3Zn

Given mass of Al = 5.00 Kg = 5000 g

Moles of Al = 5000 / 27 = 185.2 mol

From the balanced equation, it is clear that 2 mol of Al produce 3 mol of Zn.

So, 185.2 mol produce (3 x 185.2) x 2 = 277.8 mol

Moles of Zn = 277.8 mol.

MAss of Zn = 277.8mol x 65.37 g/mol = 18159.789 g = 18.2 Kg

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4.

Balancing the equation,

C3H8 + 5O2 ----------> 3CO2 + 4H2O

Given, mass of CO2 = 300. g

Moles of CO2 = 300 / 44 = 6.82 mol.

From the balanced equation, it is clear that 3 mol of CO2 is produced from 1 mol propane.

So, 6.82 mol CO2 is produced from (6.82 / 3) = 2.27 mol propane.

Moles of propane = 2.27

Mass of propane = 2.27 mol x 44 = 100. g of propane.

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5.

Balancing the equation

2H2 + O2 ----------> 2H2O

a.

Given, mass of each reactant = 45 mg = 0.045 g

Moles of H2 = 0.045 / 2 = 0.0225

Moles of O2 = 0.045 / 32 = 1.406 x 10^-3

In order to find the limiting reagent, the moles of each reactant are divided by their respective stoichiometric coefficients.

So, moles of H2 = 0.0225 / 2 = 0.01125

Moles of O2 = 0.001406

As the moles of O2 are less in number, O2 is the limiting reagent.

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b.

From the balanced equation, 1 mol of O2 produces 2 mol of H2O.

So, 1.406 x 10^-3 mol of O2 produces 2.812 x 10^-3 mol of H2O

Mass of water = 2.812 x 10^-3 x 18 = 0.051 g

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6.

Percent yield = (Practical yield / Theoretical yield) x 100

Limiting reagent = It is the reactant which limits the formation of the product, and consumed the earliest in the reaction.

Excess reagent = It is the reactant which is present in excess, and remains after the reaction is complete.

--------------------

7.

Percent yield = (Practical yield / Theoretical yield) x 100

Predicted yield = Theoretical yield = 45.8 g

Actual amount collected = Practical yield = 37.5 g

Percent yield = (37.5 / 45.8 ) x 100 = 81.8 %

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