What mass of oxygen gas can be be produced when 1.15 g sodium peroxide reacts with excess water?

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

2 Answers
Jul 3, 2015

Answer:

The mass of oxygen that be produced from 1.15 g sodium peroxide is 0.236 g.

Explanation:

#"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 #"1.15 g"# using its molar mass.

#1.15color(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)))="0.014748 mol Na"_2"O"_2"#

Determine the number of moles of #"O"_2"# that can be produced by #"0.014748 mol Na"_2"O"_2# by multiplying the moles of #"Na"_2"O"_2# times the mole ratio with #"O"_2# in the numerator.

#0.014748 color(red)cancel(color(black)("mol Na"_2"O"_2))xx("1 mol O"_2)/(2 color(red)cancel(color(black)("mol Na"_2"O"_2)))="0.0073740 mol O"_2"#

Determine the mass in grams in #"0.0073740 mol O"_2"# by multiplying the moles #"O"_2"# times its molar mass.

#0.0073740 color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="0.236 g O"_2"#

Jul 4, 2015

Answer:

First balance the equation to find that so the moles of #O_2# released is equal to one-half the moles of #Na_2O_2# added:
#2 Na_2O_2 + 2 H_2O rarr 4 NaOH + O_2#

Explanation:

The molar mass of #Na_2O_2# is 77.98 g/mol, so the moles of #Na_2O_2# is
#(1.15g)/(77.98 g/(mol))=0.01456# mol
(carries 1 extra sig fig for intermediate calculation)

The molar mass of #O_2# is 32.00 g/mol, so the mass of #O_2# released is
#1/2 times 0.01456 mol times 32.00 g/(mol)=0.233#g