When table sugar (sucrose, C₁₂H₂₂O₁₁) is heated, it decomposes to form C and H₂O.
b. How many grams of carbon are formed by the breakdown of 60.0 g of sucrose?

Question

When table sugar (sucrose, C₁₂H₂₂O₁₁) is heated, it decomposes to form C and H₂O.

b. How many grams of carbon are formed by the breakdown of 60.0 g of sucrose?

Accepted Answer

Welcome back everyone. When octane C eight H 18 is burned in a very limited amount of oxygen, the result is the formation of carbon and water. Calculate the mass and grams of carbon formed from the partial burning of one liter of octane. And note that the density of octane is 0.6 99 g per cubic centimeter. Let's begin with our first step which is to write out our equation where we have octane C eight or sorry, that should say C eight H 18, we're told that we're reacting with a limited amount of oxygen. So we have 02 as our second reactant. Now recall that octane is a gas. So it's a fuel. So it gets the liquid face and oxygen is in its natural form of gas. As our products, we have carbon formed, which is a solid and then we have water, which is a liquid. So those are two products. And now with this reaction written, we want to make sure that this is balanced by comparing our reactant to our products. So on the reactant side note that we have eight moles of carbon, we have 18 moles of hydrogen and we have two moles of oxygen. On our product side, we can count one mole of carbon to moles of hydrogen and one mole of oxygen recall that we should begin by balancing non non hydrogen and non oxygen atoms first. So focusing on our carbon atoms, we want to think of a multiple between our eight moles of carbon on the reactive side and our one mole of carbon on the product side, that would allow these two To be equal on both sides. So because we have eight moles of carbon on the reactant side, if we place a coefficient of two in front of our octane, that would now change our eight moles of carbon on the reacting side to 16 moles of carbon where we could easily place a coefficient of 16 in front of carbon on the product side to also give us 16 moles of carbon. Now notice that we've changed our number of hydrogen on the reactant side by placing the two in front of our octane. So this would give us 18 times two for hydrogen, which would give us 36 moles of hydrogen in which we need to bounce that out on the product side. So going back to our product side, we have two moles of hydrogen And we're going to be able to place a coefficient of now 18 on the product side in front of water, which will change our number of hydrogen to 36 on the product side. So our hydrogen is balanced and that will change our oxygen on the product side to also 18 being 18 times one since oxygen does not have a subscript. So now on the reactant side, we're going to be able to place a coefficient of nine in front of oxygen gas on the reactant side. And multiplying by the subscript of two would give us 18 moles of oxygen on the reactant side. Meaning that we now have a balanced equation note that we're given our volume of octane, which is our reactant and as well as its density. Now, we are missing its molar mass. And let's recall that the molar mass of octane C eight H as found from taking our eight moles of carbon multiplied by carbons molar mass from the periodic table of 12 point oh one g per mole. This is then added to our 18 moles of hydrogen multiplied by hydrogen molar mass from the periodic table of 1. grams per mole. So taking the sum here, we would find a molar mass of octane equal to 118.2, 24 grams per mole. So we're going to keep this information to the side. And for our next step, we're going to begin with our volume of octane given as 1. leaders of CAH- 18 octane. So from our volume of octane, we want to get to our given density. So we want to cancel out our units of leaders to get two cubic centimeters. So we're going to multiply by our first conversion factor in which we should recall that one leader in the denominator has an equivalent of one cubic meter. So canceling out leaders, we have cubic estimators in our numerator. And now we want to multiply by our next conversion factor to go from one decimate er in the denominator, which has an equivalence of one times 10 to the negative first power meters in the numerator. And we would need this to be in cubic meters. So we're going to take this conversion factor and raise it to an exponent outside of parentheses of three. And so this allows us to cancel our units of cubic decima eaters leaving us with units of cubic meters. And so next, we're going to multiply by our next conversion factor where we recall that one m in the denominator has an equivalent of 10 to the second power centimeters in the numerator. And because we need cubic meters, we're going to place this this conversion factor outside of parentheses to an exponent of three. So now canceling out cubic meters, we have cubic centimeters in our numerator in which we can now incorporate are given density of octane as 30.6 grams in the numerator. And this is per cubic centimeter in the denominator. And so canceling out cubic centimeters, we now have units of grams in the numerator in which we will now incorporate our molar mass of octane as our multiplier. Next, where we have an equivalence for one mole of octane in the numerator C eight h 18 In our denominator, this is equal to its molar mass of 100 and 14.2 to 4 g. And just to make a correction, the molar mass I should have written was 114.2 to 4 g per mole. So, apologies for this mistake. So now notice that we can cancel our units of grams. We have moles of octane, but we need molds of our product in which. Now we're going to refer to our balanced equation to look at our ratio between octane two carbon, Our product. And so we would see from our coefficients, we have a 2-16 molar ratio. So as our next multiplier, we want to multiply to go from two moles of octane in the denominator and will actually carry this below so that we have enough room. So continuing our stock geometry, we have two moles of C eight H 18 and the denominator, We're in our numerator, we have an equivalent of 16 moles of carbon. Now going back to our prompt, it states that we need to calculate the mass and grams of carbon formed. So we're going to need to cancel our units of moles of carbon. But first, let's cancel our moles of octane. So now from moles of carbon, we're going to multiply by our last conversion factor to go from one mole of carbon in the denominator. We're in our numerator, we plug in the molar mass of carbon which has an equivalent of 12.01 g of carbon. And this is Permal as we have in our denominator. So now canceling out our units of moles of carbon were left with grams of carbon in our numerator as our final units. And this is going to result in our final answer of 587. g of carbon that is produced. So rounding off to our minimum of 36 fix, we would round this to about 588 g of carbon produced from the partial burning of one liter of octane. So 588 g of carbon produced as our final answer corresponding to choice d in the multiple choice. I hope this made sense and let us know if you have any questions.

When table sugar (sucrose, C₁₂H₂₂O₁₁) is heated, it decomposes to form C and H₂O.
b. How many grams of carbon are formed by the breakdown of 60.0 g of sucrose?

Verified video answer for a similar problem:

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When table sugar (sucrose, C12H22O11) is heated, it decomposes to form C and H2O. b. How many grams of carbon are formed by the breakdown of 60.0 g of sucrose?

Verified video answer for a similar problem:

Key Concepts

Chemical Decomposition

Stoichiometry

Molar Mass

When table sugar (sucrose, C₁₂H₂₂O₁₁) is heated, it decomposes to form C and H₂O.
b. How many grams of carbon are formed by the breakdown of 60.0 g of sucrose?