rams Part A What is the molar mass of butane, CH10? Express your answer to four

Question

rams Part A What is the molar mass of butane, CH10? Express your answer to four significant figures and include the appropriate units. View Available Hint(s) 58.12-5 mole Correct The molar mass of C4Hio is based on the number of each kind of atom in the formula, and values from the periodic table the nmolar massesfor li o an200/moe)+ 10(1.008 g/mole)-58.12g/mole H0: 2(1.008 g/mole) + (16.00 g/mole) 18.02 g/mole CO (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 to starts 1. 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 3 Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y Part E Calculate the mass of water produced when 5.00 g of butane reacts with excess oxygen Express your'answer to three significant figures and include the appropriate units. View Available Hint(s) Value Units Submit

Explanation / Answer

Answer:

Part A: The molar mass of C4H10 =58.12 g/mol, CO2=44.01 g/mol and H2O=18.02 g/mol.

Part B:

The balanced equation for combustion of butane with excess oxygen is

C4H10 +(13/2) O2 --------> 4CO2 + 5H2O

Given mass of butane=5 g

Moles of butane=mass/molar mass=5g/58.12 g/mol

Moles of butane=0.086 mol.

From the balanced equation,

Moles of H2O=5 x mol of butane=5x0.086 mol

Moles of H2O produced=0.43 mol.

The mass of H2O produced=mol x molar mass

=0.43 molx18.02 g/mol=7.75 g.

Mass of H2O produced=7.75 g.

Part C:

Given mass of CO2=54.1 g

Moles of CO2=54.1 g/44.01 g/mol=1.229 mol.

From the equation moles if C4H6=(1/4) x mol CO2

=(1/4)x1.229 mol=0.3073 mol.

Mass of butane needed=0.3073 molx58.12 g/mol= 17.86 g.

Mass of butane needed=17.86 g.

Here Part A is correct. Thanks and I hope you like it.

Please let me know if you have any doubt.

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