Question #5b51f

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Question #5b51f

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Answer 1 · 2018-01-24T22:29:51

Here's one way to do it.

Explanation:

Problem:

Assume you are reacting 20.0 g of carbon and 20.0 g of oxygen to form carbon dioxide. What is the mass of excess reactant used and what mass of excess reactant is unused?

Solution:

$M_{r} : m m m l l 12.01 m m 32.00$ $M_r:color(white)(mmmll) 12.01 color(white)(mm)32.00$
$m m m m m m m C m + m O_{2} m \to m {CO}_{2}$ $color(white)(mmmmmmm) "C" color(white)(m)+ color(white)(m)"O"_2color(white)(m) →color(white)(m) "CO"_2$
$Mass/g: m m m 20 m m m l 20$ $"Mass/g:"color(white)(mmm)20color(white)(mmml)20$
$Amt/mol: m l l 1.67 m m 0.625$ $"Amt/mol:"color(white)(mll)1.67color(white)(mm)0.625$

You calculated that:

The amount of $O_{2}$ $"O"_2$ is 0.625 mol
The amount of $C$ $"C"$ is 1.67 mol
$O_{2}$ $"O"_2$ is the limiting reactant
$C$ $"C"$ is the excess reactant

Then

$"Moles of C used" = 0.625 color(red)(cancel(color(black)("mol O"_2))) × "1 mol C"/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.625 mol C"$

$"Mass of C used" = 0.625 color(red)(cancel(color(black)("mol C"))) × "12.01 g C"/(1 color(red)(cancel(color(black)("mol C")))) = "12.5 g C"$

$"Mass of unused C = 20.0 g - 12.5 g = 7.5 g"$

Therefore,

Mass of excess reactant used = 12.5 g
Mass of unused excess reactant = 7.5 g

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Question #5b51f

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