17.2 grams #MgO# are produced according to the following equation: #2Mg + O_2 -> 2MgO#. How many grams of oxygen #O_2# must have been consumed?

1 Answer
Dec 13, 2015

Answer:

The amount of #"O"_2"# consumed is 6.83 g.

Explanation:

Balanced Equation

#"2Mg(s)" + "O"_2("g")##rarr##"2MgO(s)"#

You need the molar masses of #"MgO"# and #"O"_2"#.

Molar Masses
#"MgO":##"40.304400 g/mol"#
http://www.ncbi.nlm.nih.gov/pccompound?term=%22Magnesium+oxide+(MgO)%22

#"O"_2:##"31.9988 g/mol"#
http://pubchem.ncbi.nlm.nih.gov/compound/977#section=Top

Determine the moles of MgO by dividing the mass of MgO produced by its molar mass.

#17.2cancel"g MgO"xx(1"mol MgO")/(40.304400cancel"g MgO")="0.42675 mol MgO"#

I am keeping some guard digits to reduce rounding errors. I will round to three significant figures at the end.

Multiply the moles of MgO times the mole ratio between #"O"_2"# and #"MgO"# from the balanced equation. This will give the moles of #"O"_2"# that were consumed.

#0.42675cancel"mol MgO"xx(1"mol O"_2)/(2cancel"mol MgO")="0.21338 mol O"_2"#

Multiply the moles of #"O"_2"# times the molar mass of #"O"_2"#. This will give the mass of #"O"_2"# consumed.

#0.21338cancel"mol O"_2xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2"#

You can combine all of the steps as shown below.

#17.2cancel"g MgO"xx(1cancel"mol MgO")/(40.304400cancel"g MgO")xx(1cancel"mol O"_2)/(2cancel"mol MgO")xx(31.9988"g O"_2)/(1cancel"mol O"_2)="6.83 g O"_2"#