# How much "Ca" is consumed when "56.8 mL O"_2" reacts with calcium to produce "CaO"?

## $\text{2Ca+O"_2}$$\rightarrow$$\text{2CaO}$

##### 2 Answers
Jun 21, 2015

This reaction will consume 0.201 g of calcium.

#### Explanation:

Start by taking a look at the balanced chemical equation for this reaction

$\textcolor{red}{2} C {a}_{\left(s\right)} + {O}_{2 \left(g\right)} \to 2 C a {O}_{\left(s\right)}$

Notice that you have a $\textcolor{red}{2} : 1$ mole ratio between calcium and oxygen. This means that, regardless of how many moles of oxygen reacted, the reaction needed twice as many moles of calcium.

Since you're at STP conditions, you can use the molar volume of a gas at STP. More specifically, you know that at STP conditions, which imply a pressure of 100 kPa and a temperature of 273.15 K, 1 mole of any ideal gas occupies exactly 22.7 L.

This means that you can determine how many moles of oxygen reacted by using its volume

56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles" ${O}_{2}$

This means that the reaction also consumed

0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles" $C a$

To get the mass of calcium that contained this many moles, use calcium's molar mass

0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g" $C a$

Rounded to three sig figs, the number of sig figs you gave for the volume of oxygen, the answer will be

${m}_{C a} = \textcolor{g r e e n}{\text{0.201 g}}$

SIDE NOTE Many textbooks and online sources still list STP conditions as a pressure of 1 atm and temperature of 273.15, which would make the molar volume of a gas equal to 22.4 L, not 22.7 L.

If this is the value you're supposed to use, simply redo all the calculations using 22.4 L instead of 22.7 L. The answer should come out to be 0.203 g of calcium.

Jun 21, 2015

$0.203 \text{g Ca}$ are consumed when $56.8 {\text{mL O}}_{2}$ reacts with calcium to produce calcium oxide, $\text{CaO}$.

#### Explanation:

Balanced Equation

${\text{2Ca+O}}_{2}$$\rightarrow$$\text{2CaO}$

The mole ratio of $\text{Ca}$ to ${\text{O}}_{2}$ is $\left(2 \text{mol Ca")/(1 "mol O"_2}\right)$.

In order to solve this problem, we need to convert the volume of oxygen to liters. Then convert the volume to moles using the molar volume of a gas. Next we will multiply the moles of oxygen times the mole ratio from the equation to get moles of calcium. Finally, we will multiply the moles of calcium times its molar mass to get the mass of calcium consumed.

Convert volume to liters.

56.8color(red)cancel(color(black)("mL O"_2))xx("1L O"_2)/(1000color(red)cancel(color(black)("mL O"_2")))$= 0.0568 \text{mol O"_2}$

Convert volume in liters to moles.

The volume of one mole of a gas is its molar volume. At STP the molar volume of an ideal gas is $22.414 \text{L/mol}$.

0.0568color(red)cancel(color(black)("L O"_2))xx(1"mol O"_2)/(22.414color(red)cancel(color(black)("L O"_2)))=0.0025341"mol O"_2

Determine the moles of Ca consumed.

Multiply the moles of ${\text{O}}_{2}$ times the previously determined mole ratio between calcium and oxygen.

0.0025341color(red)cancel(color(black)("mol O"_2))xx("2mol Ca")/(1color(red)cancel(color(black)("mol O"_2)))=0.0050682"mol Ca"

Convert moles of Ca consumed to mass of Ca consumed.

We need the molar mass of $\text{Ca}$, which is $40.078 \text{g/mol}$. (This is its atomic weight on the periodic table in grams/mole.)

0.0050682color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))=0.203"g Ca" (rounded to three significant figures)