What mass of oxygen will be produced from the reaction between #"6.95 g"# of sodium peroxide #("NaO"_2")# and water?

#"2Na"_2"O"_2 + "2H"_2"O"##rarr##"4NaOH + O"_2"#

1 Answer
Jul 3, 2015

#"1.41 g O"_2"# will be produced from #"6.95 g NaO"_2"#.

Explanation:

Balanced equation.

#"2Na"_2"O"_2"+2H"_2"O"##rarr##"4NaOH"+"O"_2"#

Determine the mole ratios between #"Na"_2"O"_2# and #"O"_2"# from the balanced equation.

#("2 mol Na"_2"O"_2)/("1 mol O"_2")# and #("1 mol O"_2)/("2 mol Na"_2"O"_2")#

Determine the molar masses of #"Na"_2"O"_2# and #"O"_2#.

#"Na"_2"O"_2:##(2xx22.990)+(2xx15.999)="77.978 g/mol"#

#"O"_2:##(2xx15.999)="31.998 g/mol"#

Determine the number of moles of #"Na"_2"O"_2# in #"6.85 g"# using its molar mass.

#6.85color(red)cancel(color(black)("g Na"_2"O"_2))xx(1"mol Na"_2"O"_2)/(77.978color(red)cancel(color(black)("g Na"_2"O"_2)))="0.08785 mol Na"_2"O"_2"#

Determine the number of moles of #"O"_2"# that can be produced by #"0.08785 mol Na"_2"O"_2# by multiplying the moles of #"Na"_2"O"_2# times the mole ratio with #"O"_2# in the numerator.

#0.08785color(red)cancel(color(black)("mol Na"_2"O"_2))xx("1 mol O"_2)/(2color(red)cancel(color(black)("mol Na"_2"O"_2)))="0.04393 mol O"_2"#

Determine the mass in grams in #"0.04393 mol O"_2"# by multiplying the moles #"O"_2"# times its molar mass.

#0.04393color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="1.41 g O"_2"# (rounded to three significant figures)