CH 09 HW ± Masses of Reactants and Products ResourcesConstantsPeriodic Table « p

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CH 09 HW

± Masses of Reactants and Products

ResourcesConstantsPeriodic Table

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± Masses of Reactants and Products

2C4H10(g)+13O2(g)10H2O(g)+8CO2(g)

Also, notice that the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce.

Molar mass

The first step of many stoichiometry problems is to convert the given value from grams to moles. Molar masses, which can be found using the periodic table, serve as conversion factors between grams and moles.

Part A

What is the molar mass of butane, C4H10?

Express your answer to two decimal places and include the appropriate units.

Hints

C4H10:    4(12.01 g/mole)+10(1.008 g/mole)=58.12 g/mole

H2O:    2(1.008 g/mole)+(16.00 g/mole)=18.02 g/moleCO2:    (12.01 g/mole)+2(16.00 g/mole)=44.01 g/mole

Solving stoichiometry problems that involve mass

Convert from grams of compound X to moles of compound X using the molar mass of compound X.

Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation.

Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y.

Part B

Calculate the mass of water produced when 7.35 g of butane reacts with excess oxygen.

Express your answer to three digits and include the appropriate units.

Hints

Hint 1. Convert the mass of butane to moles

The molar mass of butane, C4H10, is 58.12 g/mole . Convert 7.35 g of butane to moles.

Express your answer to three digits and include the appropriate units.

Hint 1. Identify how to convert from grams to moles(click to open)

This hint will be shown after you complete previous hint(s).

Hint 2.

Part C

Calculate the mass of butane needed to produce 68.3 g of carbon dioxide.

Express your answer to three digits and include the appropriate units.

± Masses of Reactants and Products

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:

2C4H10(g)+13O2(g)10H2O(g)+8CO2(g)

The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react.

Also, notice that the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce.

But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios.

Molar mass

The first step of many stoichiometry problems is to convert the given value from grams to moles. Molar masses, which can be found using the periodic table, serve as conversion factors between grams and moles.

Part A

What is the molar mass of butane, C4H10?

Express your answer to two decimal places and include the appropriate units.

Hints

The molar mass of C4H10 is based on the number of each kind of atom in the formula, and values from the periodic table:

C4H10:    4(12.01 g/mole)+10(1.008 g/mole)=58.12 g/mole

Similarly, the molar masses for H2O and CO2 are

H2O:    2(1.008 g/mole)+(16.00 g/mole)=18.02 g/moleCO2:    (12.01 g/mole)+2(16.00 g/mole)=44.01 g/mole

The molar mass can be used to convert between grams and moles of a compound.

Solving stoichiometry problems that involve mass

Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is

Convert from grams of compound X to moles of compound X using the molar mass of compound X.

Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation.

Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y.

Part B

Calculate the mass of water produced when 7.35 g of butane reacts with excess oxygen.

Express your answer to three digits and include the appropriate units.

Hints

Hint 1. Convert the mass of butane to moles

The molar mass of butane, C4H10, is 58.12 g/mole . Convert 7.35 g of butane to moles.

Express your answer to three digits and include the appropriate units.

Hint 1. Identify how to convert from grams to moles(click to open)

This hint will be shown after you complete previous hint(s).

Hint 2.

Part C

Calculate the mass of butane needed to produce 68.3 g of carbon dioxide.

Express your answer to three digits and include the appropriate units.

Explanation / Answer

2C4H10(g)+13O2(g)10H2O(g)+8CO2(g)

molar mass of C4H10    = 4(12.01 g/mole)+10(1.008 g/mole)=58.12 g/mole

molar mass of H2O = 2(1.008 g/mole)+(16.00 g/mole)=18.02 g/mole

molar mass of CO2 = (12.01 g/mole)+2(16.00 g/mole)=44.01 g/mole

part-B

2C4H10(g)+13O2(g)10H2O(g)+8CO2(g)

2 moles of C4H10 react with O2 to gives 10 moles of H2O

2*58.12 g of C4H10 react with O2 to gives 10 *18.02 g of H2O

7.35 g of C4H10 react with O2 to gives = 10*18.02*7.35/2*58.12    = 11.4g of H2O

part-C

2C4H10(g)+13O2(g)10H2O(g)+8CO2(g)

8 moles of CO2 produced from 2 moles of C4H10

8*44.01g of CO2 produced from 2*58.12 g of C4H10

68.3g of CO2 produced from = 2*58.12*68.3/8*44.01   =22.6g of C4H10   >>>>answer

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