Question #d4a87
1 Answer
Here's what I got.
Explanation:
Notice that according to the balanced chemical equation
#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#
the reaction consumes equal numbers of moles of carbon and oxygen gas. Moreover, the number of moles of carbon dioxide produced is equal to the numbers of moles of the two reactants consumed.
Your goal now is to to convert the mass of carbon first from kilograms to grams by using the conversion factor
#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))#
then from grams to moles by using carbon's molar mass, i.e. the mass of One mole of carbon, which is equal to
So, the mass of carbon given to you will be equivalent to
#"12,700,586" color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.057413 * 10^9"moles C"#
Since you know that the reaction consumes equal number of moles of carbon and of oxygen gas, it follows that this much carbon will require
#"moles of O"_2= 1.057413 * 10^9"moles O"_2#
To convert this to grams, use oxygen gas' molar mass, which is is equal to
#1.057413 * 10^9color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(3.3837216 * 10^10"g O"_2)color(white)(a/a)|)))#
To find the mass of carbon dioxide produced by the reaction, use the fact that the reaction produces
#n_(CO_2) = 1.057413 * 10^9"moles CO"_2#
and carbon dioxide's molar mass, which is equal to
#1.057413 * 10^9 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)(4.6526172 * 10^(10)"g CO"_2)color(white)(a/a)|)))#
I'll leave the answers rounded to eight sig figs, since that's how many sig figs you have for the mass of carbon.
