The equation for burning C2H2 is
2C2H2(g) + 5O2(g) ⎯→ 4CO2(g) + 2H2O(g)
a. If 15.9 L C2H2 react at STP, how many
moles of CO2 are produced? (Hint: At
STP, 1 mol = 22.41 L for any gas.)
b. How many milliliters of CO2 (density =
1.977 g/L) can be made when 59.3 mL O2
(density = 1.429 g/L) react?
2.39
Write and balance the proper reaction for ethyne (C2H2) being burned in an oxygen atmosphere.
a. If 15.9 L C2H2 react at standard temperature and pressure, how many moles of CO2 are produced?
b. How many milliliters of CO2 (density = 1.977 g/L) can be made when 59.3 mL O2 (density = 1.429 g/L) react?
a. Well, if 1 mole of C2H2 produces 4 moles of CO2, we need to find out how many moles of C2H2 we have.
Using the equation:
1 mol C2H2 = 22.41 L C2H2,
15.9 L C2H2 = (15.9/22.41) mol C2H2.
Now, since the stoichiometry of the equation is 2:4 for C2H2:CO2, we can say that 15.9 L C2H2 will produce (4/2) * (15.9/22.41) = (2/2.812) = 0.710 moles of CO2.
b. First, let's find out how many grams of O2 are present:
Using the density of O2,
59.3 mL O2 = (59.3/1000) L O2,
Mass of O2 = volume * density = (59.3/1000) * 1.429 g = 0.0847 g O2.
Now, we need to find the number of moles of O2 using its molar mass:
1 mole of O2 = 32.00 g O2,
0.0847 g O2 = (0.0847/32.00) mol O2 = 0.002646 mol O2.
Looking at the equation, we see that the stoichiometry is 5:4 for O2:CO2. This means that 0.002646 mol O2 will produce (4/5) * 0.002646 = 0.002117 mol CO2.
To find the volume of CO2 produced, we use the ideal gas law:
1 mol CO2 = 22.41 L CO2,
0.002117 mol CO2 = (0.002117 * 22.41) L CO2 = 0.0474 L CO2.
So, the volume of CO2 produced is 0.0474 L, but since we want the answer in milliliters, that's 0.0474 * 1000 = 47.4 mL. Phew, that was quite a journey, wasn't it? Punny calculations!
a. To calculate the number of moles of CO2 produced when 15.9 L of C2H2 react at STP, we can use the given hint: "At STP, 1 mol = 22.41 L for any gas."
Step 1: Convert the volume of C2H2 from liters to moles.
Given: Volume of C2H2 = 15.9 L
We can use the conversion factor 1 mol C2H2 = 22.41 L C2H2 (from the hint).
moles of C2H2 = (15.9 L C2H2) / (22.41 L C2H2/mol C2H2)
Step 2: Use the stoichiometry of the balanced equation to find the moles of CO2 produced.
From the balanced equation, the mole ratio of C2H2 to CO2 is 2:4.
moles of CO2 = (moles of C2H2) * (4 mol CO2 / 2 mol C2H2 )
b. To calculate the number of milliliters of CO2 that can be produced when 59.3 mL of O2 reacts, we need to use the densities of O2 and CO2 to convert from volume to mass.
Step 1: Convert the volume of O2 from milliliters to grams.
Given: Volume of O2 = 59.3 mL
Density of O2 = 1.429 g/L
mass of O2 = (volume of O2) * (density of O2)
Step 2: Convert the mass of O2 to moles using the molar mass.
The molar mass of O2 is approximately 32 g/mol. Divide the mass of O2 by its molar mass to get the moles of O2.
moles of O2 = (mass of O2) / (molar mass of O2)
Step 3: Use the stoichiometry of the balanced equation to find the moles of CO2 produced.
From the balanced equation, the mole ratio of O2 to CO2 is 5:4.
moles of CO2 = (moles of O2) * (4 mol CO2 / 5 mol O2)
Step 4: Convert the moles of CO2 to volume using the conversion factor from the hint:
1 mol CO2 = 22.41 L CO2
volume of CO2 = (moles of CO2) * (22.41 L CO2 / 1 mol CO2)
Step 5: Convert the volume of CO2 from liters to milliliters (mL) if necessary.
a)
Convert 15.9 L C2H2 to moles. moles = L/22.41.
Convert moles C2H2 to moles CO2 using the coefficients in the balanced equation.
b)
Convert 59.3 mL O2 to moles, convert moles O2 to moles CO2, convert moles CO2 to mL.
Post your work if you get stuck.