If 220. mL of 0.1364 M aqueous LiOH and 0.3553 L of gaseous CO2 measured at 1.36 atm pressure and 360 K are reacted stoichiometrically according to the equation, what mass (g) of gaseous CO2 remained?
2 LiOH(aq) + CO2(g) → Li2CO3(aq)
moles LiOH = M x L.
Using the coefficients in the balanced equation, convert moles LiOH to moles CO2 used in the reaction.
Now use PV = nRT to calculate moles CO2 initially, subtract moles CO2 used (from above), then convert moles CO2 remaining to grams. g = moles x molar mass.
Post your work if you get stuck.
To find the mass of gaseous CO2 that remains, we need to determine the amount of CO2 that reacts and compare it to the initial amount.
First, let's find the moles of LiOH and CO2 using the given volumes and molarity:
1. Convert the volume of LiOH to liters: 220 mL = 0.220 L
2. Calculate the number of moles of LiOH:
Moles of LiOH = volume (L) x concentration (M)
= 0.220 L x 0.1364 M
3. Convert the volume of CO2 to moles using the ideal gas law:
PV = nRT (Pressure x Volume = moles x Gas constant x Temperature)
Rearrange the equation to solve for moles (n):
n = PV/RT
n = (1.36 atm x 0.3553 L) / (0.0821 atm·L/mol·K x 360 K)
Now that we have the moles of LiOH and CO2, we can determine the stoichiometric ratio between them based on the balanced equation:
2 LiOH(aq) + CO2(g) → Li2CO3(aq)
The ratio is 2:1, which means 2 moles of LiOH react with 1 mole of CO2. We need to check if we have an excess of LiOH or CO2.
If the moles of LiOH are greater than twice the moles of CO2, then all of the CO2 will have reacted, and none of it remains. If the moles of LiOH are less than twice the moles of CO2, one mole of CO2 reacts with each mole of LiOH, and the excess of CO2 remains.
Finally, with the amount that remains, we can calculate the mass of gaseous CO2 in grams by using its molar mass.
I hope this explanation helps you understand how to solve the problem!