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

Question

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

Olivia Anton · Accepted Answer

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

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"#

Owen Ambrose · Answer

undefined To determine the number of moles of $ \text{Ag}_2\text{O} $ needed to produce $ 4.24 \, \text{mol} $ of $ \text{O}_2 $, we need to use the stoichiometry of the reaction involving the conversion of $ \text{Ag}_2\text{O} $ to $ \text{O}_2 $. From the balanced chemical equation, we can see that $ 2 $ moles of $ \text{Ag}_2\text{O} $ produce $ 1 $ mole of $ \text{O}_2 $. Therefore, to find the number of moles of $ \text{Ag}_2\text{O} $ needed to produce $ 4.24 \, \text{mol} $ of $ \text{O}_2 $, we use the ratio:

$$
\text{Moles of } \text{Ag}_2\text{O} = \frac{4.24 \, \text{mol} \, \text{O}_2}{1} \times \frac{2 \, \text{mol} \, \text{Ag}_2\text{O}}{1 \, \text{mol} \, \text{O}_2} = 8.48 \, \text{mol} \, \text{Ag}_2\text{O}
$$

So, $ 8.48 \, \text{mol} $ of $ \text{Ag}_2\text{O} $ are needed to produce $ 4.24 \, \text{mol} $ of $ \text{O}_2 $.