When a mixture of 10.0 g of acetylene 1C2H22 and 10.0 g of
oxygen...

Question

When a mixture of 10.0 g of acetylene 1C2H22 and 10.0 g of
oxygen 1O22 is ignited, the resulting combustion reaction
produces CO2 and H2O. 
(c) How many grams of CO2 are present after the reaction is complete?

Accepted Answer

Welcome back everyone in this example, we're told that ethanol is a fuel made from renewable sources and produces less carbon dioxide than traditional fossil fuels. We need to calculate what massive carbon dioxide gas is produced when 13.5 g of ethanol undergoes complete combustion with 45 g of oxygen. So we want to recall that in a combustion reaction, we have our re agent which in this case is ethanol, which we recall is the formula C two H five oh H. It's reacting with oxygen and in combustion reactions, our products are always carbon dioxide and water. And as far as our labels are, ethanol is going to be a liquid and the rest of our re agents are going to have the gaseous label. Our next step is to ensure that this equation is balanced. And when we look at the molds of each of our Atoms on both sides of our equation, we're going to have to place a coefficient of three in front of our 02, a coefficient of two in front of our CO2 and a coefficient of three in front of our water so that we can ensure we have a balanced equation. Now, the prompt is asking us to figure out the mass of carbon dioxide that is produced is essentially asking us for our theoretical yield of our carbon dioxide. So we're going to need to find the theoretical yield of carbon dioxide based on both of our reactant. So, first we're going to find the theoretical yields for the mass of carbon dioxide that is produced. And sorry, that's CO2. So for the massive carbon dioxide that is produced from ethanol. So we want to make note of our molar ratio between our ethanol And our carbon dioxide. And we would get that from our balanced equation where we see we have a coefficient of one in front of ethanol and then we have a coefficient of two in front of our carbon dioxide. So we have a 1-2 molar ratio. So we're going to begin our calculation with our massive ethanol from the prompt given us 13.5 g of ethanol and we want to cancel out our units of grams of ethanol to get two moles. And so we're going to recall our molar mass of ethanol from the periodic table which we see as equal to 46.7 g of ethanol For one mole of ethanol. So now we're able to cancel out our units of grams of ethanol. And now we want to focus on going from moles of ethanol to moles of our product carbon dioxide. And we would get that from our molar ratio, which we noted to the left where we said we have one mole of ethanol for one mole of carbon dioxide. So now we're able to cancel out our moles of ethanol. And now we want to get two g of carbon dioxide. So we would recall our molar mass of carbon dioxide from the periodic table, which we understand for one mole of carbon dioxide is 43.99 g of CO2. So now we can cancel out our units of moles of carbon dioxide And we're left with our final unit being grams of carbon dioxide. And this is going to produce a mass from our ethanol of our carbon dioxide being 25.8 g of CO2. So now we want to compare this to the theoretical yields for our mass of carbon dioxide from our second reactant being oxygen. So making note of our molar ratio between oxygen and carbon dioxide, we see from our balanced equation that that ratio is a two 23 Molar ratio. And so beginning with our mass of oxygen from the prompt, we have a mass of 45.0 g of oxygen. We're going to multiply to get two moles of oxygen by recalling our molar mass from the product table, which we understand is 32 g of oxygen for one mole of oxygen. So now we can cancel out our grams of oxygen. And now we want to get from moles of oxygen. Two moles of C. 02. So, using our ratio that we took note of, we said, we have two moles of oxygen for three moles of C. 02. So now we can cancel out moles of 02 and focus on going from moles of C 022 g of C 02. So we're calling from the periodic table, our molar mass of C. 02. We see that we have one mole of C. 02 for 43.99 grams of C. 02. And now we can cancel our moles of C. 02. We're left with grams of C. 02. And this gives us a mass equal to 41.2 g of C. 02 produced from our re agent oxygen. And because we can see that we have only 25.8 g of C. 02, which is less than 41.2 g of C. 02. We would say that therefore our ethanol which makes that lower amount of carbon dioxide. So our ethanol is our limiting reactant and the reaction the reaction produces the mass of carbon dioxide being 25.8 g of C. 02. And so for our final answer, we have that highlighted in yellow here that our mass of carbon dioxide is 25.8 g produced from our ethanol. So actually we should just highlight the mass of C. 02 produced. So what's highlighted in yellow is our final answer. I hope that everything I explained was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video

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