How many moles #"Ag"_2"O"# are needed to produce #"4.24 mol O"_2"#?

#"2Ag"_2"O"##rarr##"4Ag + O"_2"#

1 Answer
May 21, 2016

Answer:

#"8.50 mol Ag"_2"O"# are needed to produce #"4.25 mol O"_2"#.

Explanation:

Balanced Equation

#"2Ag"_2"O"##rarr##"4Ag"+"O"_2"#

Multiply the given moles of #"O"_2"# times the mole ratio between #"Ag"_2"O"# and #"O"_2"# from the balanced equation, with #"Ag"_2"O"# as the numerator.

#4.25cancel"mol O"_2xx(2"mol Ag"_2"O")/(1cancel"mol O"_2)="8.50 mol Ag"_2"O"#