Question 51550

Oct 2, 2015

$V = \text{30. mL}$

Explanation:

Since it's obvious that your question is missing some important pieces of information, I will have to make some assumptions as to what is going on.

You're probably looking at the reaction between copper metal and oxygen. When copper metal is heated in the presence of oxygen it forms copper(II) oxide, or $\text{CuO}$, according to the balanced chemical equation

$\textcolor{red}{2} {\text{Cu"_text((s]) + "O"_text(2(g]) -> 2"CuO}}_{\textrm{\left(s\right]}}$

Now, copper's molar mass is equal to $\text{63.546 g/mol}$, which means that your sample of copper will contain

0.16color(red)(cancel(color(black)("g Cu"))) * "1 mole Cu"/(63.456color(red)(cancel(color(black)("g Cu")))) = "0.002518 moles Cu"

If you go by the number of sig figs you gave for the mass of copper, then the number of moles of copper will be $0.0025$.

So I assume that maybe this is what you had to calculate in the first part of the question.

This means that you would need

0.0025color(red)(cancel(color(black)("moles Cu"))) * ("1 mole O"""_2)/(color(red)(2)color(red)(cancel(color(black)("moles Cu")))) = "0.00125 moles O"""_2#

To get the volume of oxygen that would contain this many moles, you need the pressure and the temperature of the oxygen.

I would assume that you're studying this reaction at room temperature, i.e. at a temperature of ${20}^{\circ} \text{C}$ and a pressure of $\text{1 atm}$.

Use the ideal gas law equation to find the volume

$P V = N R T \implies V = \frac{n R T}{P}$

$V = \left(0.00125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 20)color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm}}}}\right)$

$V = \text{0.03005 L}$

Expressed in mililiters and rounded to two sig figs, the answer will be

$V = \textcolor{g r e e n}{\text{30. mL}}$