Question #2d295
1 Answer
Explanation:
Start by looking at the balanced chemical equation that describes this reaction
#color(blue)(2)"CO"_ ((g)) + "O"_ (2(g)) -> 2"CO"_ (2(g))#
You know that every
You already know the mass of carbon dioxide produced by the reaction, so use the compound's molar mass to convert it to moles
#1.38 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01 color(red)(cancel(color(black)("g")))) = "0.03136 moles CO"_2#
Now, you can assume that the oxygen gas is in excess because the problem doesn't mention a specific sample of this reactant.
Use the
#0.03136 color(red)(cancel(color(black)("moles CO"_2))) * (color(blue)(2)color(white)(.)"moles CO")/(2color(red)(cancel(color(black)("moles CO"_2)))) = "0.03136 moles CO"#
Finally, convert this to grams by using the compound's molar mass
#0.03136 color(red)(cancel(color(black)("moles CO"))) * "28.01 g"/(1color(red)(cancel(color(black)("mole CO")))) = color(darkgreen)(ul(color(black)("0.878 g")))#
The answer is rounded to three sig figs, the number of sig figs you have for the mass of carbon dioxide.