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

Question

#"2Ca+O"_2"##rarr##"2CaO"#

Henry Antrim · Accepted Answer

This reaction will consume 0.201 g of calcium.

Examine the balanced chemical equation for this reaction first. #color(red)(2)Ca_((s)) + O_(2(g)) -> 2CaO_((s))# Notice that you have a #color(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. You can use the molar volume of a gas at STP because you are at STP conditions. Specifically, you know that 1 mole of any ideal gas occupies exactly 22.7 L at STP conditions, which imply a pressure of 100 kPa and a temperature of 273.15 K. This implies that you can use the volume of the oxygen to calculate how many moles of oxygen reacted. #56.8cancel("mL") * (1cancel("L"))/(1000cancel("mL")) * "1 mole"/(22.7cancel("L")) = "0.002502 moles"# #O_2# This indicates that the response was also absorbed. #0.002502cancel("moles"O_2) * (color(red)(2)"moles Ca")/(1cancel("mole"O_2)) = "0.005004 moles"# #Ca# Use the molar mass of calcium to determine the mass of calcium containing this many moles. #0.005004cancel("moles") * "40.078 g"/(1cancel("mole")) = "0.2006 g"# #Ca# The answer, rounded to three sig figs, will be the number of sig figs you provided for the oxygen volume. #m_(Ca) = color(green)("0.201 g")# SIDE NOTE: A lot of textbooks and internet resources continue to state that the STP conditions are 1 atm of pressure and 273.15% of temperature. This would imply that a gas's molar volume is 22.4 L rather than 22.7 L. If this is the correct value, just repeat all the calculations with 22.4 L instead of 22.7 L. The calcium content should be 0.203 g.

Serenity Allen · Answer

#0.203"g Ca"# are consumed when #56.8"mL O"_2# reacts with calcium to produce calcium oxide, #"CaO"#.

Equilibrium Formula #"2Ca+O"_2##rarr##"2CaO"# The mole ratio of #"Ca"# to #"O"_2# is #(2 "mol Ca")/(1 "mol O"_2")#. First, we must convert the oxygen volume to liters. Next, we must use the molar volume of a gas to convert the volume to moles. Finally, we must multiply the moles of oxygen by the mole ratio from the equation to obtain moles of calcium. Lastly, we must multiply the moles of calcium by its molar mass to obtain the mass of calcium consumed. Calculate the volume in liters. #56.8color(red)cancel(color(black)("mL O"_2))xx("1L O"_2)/(1000color(red)cancel(color(black)("mL O"_2")))##=0.0568"mol O"_2"# Convert liters of volume 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"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# Count the moles of calcium that were ingested. Multiply the moles of #"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"# Calculate the mass of calcium consumed in moles. We need the molar mass of #"Ca"#, which is #40.078"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)

Peyton Adams · Answer

undefined To find out how much calcium (Ca) is consumed when 56.8 mL of oxygen (O2) reacts with calcium to produce calcium oxide (CaO), we need to first balance the chemical equation for the reaction:

2 Ca + O2 → 2 CaO

From the balanced equation, we can see that 1 mole of O2 reacts with 2 moles of Ca to produce 2 moles of CaO.

Next, we need to convert the volume of O2 to moles. We can use the ideal gas law for this:

PV = nRT

where:
P = pressure (assume 1 atm)
V = volume (convert 56.8 mL to L, so V = 0.0568 L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (assume room temperature, around 298 K)

Now we can calculate the number of moles of O2:

n(O2) = (P * V) / (R * T)
n(O2) = (1 atm * 0.0568 L) / (0.0821 L·atm/mol·K * 298 K)
n(O2) ≈ 0.0023 moles

Since 1 mole of O2 reacts with 2 moles of Ca, the number of moles of Ca consumed will be half of the moles of O2:

n(Ca) = 0.0023 moles / 2
n(Ca) = 0.00115 moles

Now, we can calculate the mass of Ca consumed using the molar mass of Ca:

m(Ca) = n(Ca) * molar mass(Ca)
m(Ca) = 0.00115 moles * 40.08 g/mol
m(Ca) ≈ 0.0462 grams

Therefore, approximately 0.0462 grams of calcium is consumed when 56.8 mL of oxygen reacts with calcium to produce calcium oxide.