How much #"Ca"# is consumed when #"56.8 mL O"_2"# reacts with calcium to produce #"CaO"#?
#"2Ca+O"_2"# #rarr# #"2CaO"#
This reaction will consume 0.201 g of calcium.
Start by taking a look at the balanced chemical equation for this reaction
Notice that you have a
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
This means that the reaction also consumed
To get the mass of calcium that contained this many moles, use calcium's molar mass
Rounded to three sig figs, the number of sig figs you gave for the volume of oxygen, the answer will be
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.
The mole ratio of
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.
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
Determine the moles of Ca consumed.
Multiply the moles of
Convert moles of Ca consumed to mass of Ca consumed.
We need the molar mass of