What volume of O2(g) is required to react with excess CS2(l) to produce 10.0L of CO2(g)? (Assume all gases are measured at 0°C and 1atm.)

Question

What volume of O2(g) is required to react with excess CS2(l) to produce 10.0L of CO2(g)? (Assume all gases are measured at 0°C and 1atm.)

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

Show all calculations and put your answer in correct sig figs.

Answer 1

Hello, Sam,

Please share what you have already done on a problem before posting. It helps guide where the explanation should focus. I'll summarize the key steps. Please ask if you don't understand a particular step.

We are fortunate that all the gases are at STP. I was already jotting down the gas law before I saw STP. Thanks to a handy feature of all gases, we have a conversion factor that tells us how many moles we have, based only on it's volume at STP. That is: all gasses occupy 22.4 liters/mole at STP. Now we can covert the 10.0 liters of CO₂ directly into moles CO₂ by dividing by 22.4 liters/mole:

(10.0 liters CO₂)/(22.4 liters CO₂/mole CO₂) = 0.446 moles CO₂

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

The balanced equation says we need 3 moles of O₂ for every 1 moles of CO₂ produced. That's a molar ratio of 3/1 (molesO₂/mole CO₂).

Find the moles O₂ required by multiplying this ratio trimes the moles of CO₂ actually produced:

(0.466 moles CO₂)*(3 molesO₂/mole CO₂) = 1.34 moles O₂

Now we're almost done. The question asks for volume of O₂ at STP, so we can use the same conversion factor we used in the forst step:

(1.34 moles O₂)*(22.4 liters O₂/mole O₂) = 30.0 liters O₂ (3 sig figs)

I hope that helps,

Bob

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Extra credit:

We can find grams O₂:

Convert the moles oxygen to grams oxygen by multiplying by it's molar mass:

(1.34 moles O₂)*(32 grams O₂/mole O₂) = 42.9 grams O₂