Butane gas (C4 H10) burns according to the following...

Butane gas $\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)$ burns according to the following equation: $(9.2,9.3,9.5)$ $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ a. How many moles of $\mathrm{H}_{2} \mathrm{O}$ are produced from the complete reaction of $2.50 \mathrm{~mol}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ? b. How many grams of $\mathrm{O}_{2}$ are needed to react completely with $22.5 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ? c. How many grams of $\mathrm{CO}_{2}$ are produced from the complete reaction of $55.0 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ? d. If the reaction in part c produces $145 \mathrm{~g}$ of $\mathrm{CO}_{2}$, what is the percent yield of $\mathrm{CO}_{2}$ for the reaction?

Butane gas $\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)$ burns according to the following equation: $(9.2,9.3,9.5)$
$$
2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)
$$
a. How many moles of $\mathrm{H}_{2} \mathrm{O}$ are produced from the complete reaction of $2.50 \mathrm{~mol}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ?
b. How many grams of $\mathrm{O}_{2}$ are needed to react completely with $22.5 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ?
c. How many grams of $\mathrm{CO}_{2}$ are produced from the complete reaction of $55.0 \mathrm{~g}$ of $\mathrm{C}_{4} \mathrm{H}_{10}$ ?
d. If the reaction in part c produces $145 \mathrm{~g}$ of $\mathrm{CO}_{2}$, what is the percent yield of $\mathrm{CO}_{2}$ for the reaction?

Key Concepts

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole, and is used to convert between the mass of a substance and the number of moles. This concept is crucial when performing stoichiometric calculations where the quantity of reactants or products is given in grams and needs to be converted to moles for further analysis.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, defined as the ratio of the actual yield (the amount of product actually obtained from the reaction) to the theoretical yield (the amount predicted by stoichiometric calculations), multiplied by 100. This concept helps in evaluating the performance of a reaction and determining whether any loss or side reactions occurred during the process.

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction with equal numbers of atoms for each element on both sides of the reaction arrow, ensuring the law of conservation of mass is observed. It provides the stoichiometric coefficients that indicate the mole ratios of reactants and products, which are critical for quantitative calculations in chemical reactions.

Mole Concept and Mole Ratio

The mole concept is a fundamental unit in chemistry that relates the number of particles in a substance to its mass (in grams). In the context of a chemical reaction, mole ratios derived from the balanced equation are used to convert between quantities of reactants and products, allowing for stoichiometric calculations such as determining the number of moles or grams of a substance involved in the reaction.

Transcript

00:01 Butane, used as a fuel in disposable lighters, reacts with oxygen as follows.

00:12 It's a big storm here.

00:28 We are asked to do a whole bunch of different things here.

00:31 First, we're asked to balance the chemical equation.

00:37 Then we're going to do some limiting reactant work.

00:44 And then we're going to do some like percent yield stuff, i think.

00:51 Maybe not.

00:52 I'm not looking at the question.

00:53 So, butane is 2, 13, 8, and 10.

01:01 So we're going to have some fun coefficients to work with here.

01:05 Next, we have 10 grams and 10 grams.

01:10 So we are given 10 .00 grams of each of these two reactants.

01:24 What is the theoretical yield of each product and how many grams of excess are left over? and how many additional grams of limiting are needed? okay.

01:38 So first, let's just pick either product.

01:40 I'll pick the co2.

01:43 I'm going to go to moles of co2 to figure out which one's limiting.

01:46 After i have the limiting, then we will figure out the mass of each product form.

01:53 Let's start with our 10 grams of our butane and our 10 grams of our oxygen.

02:27 And i would probably go 14 .14 for that, but i'm going to type my butane.

02:35 In to this.

02:40 So i've got 58 .12 grams of butane per mole of butane.

03:08 Our next step is to use our mole ratios and i've got butane in my denominator here and i said we were going to go to co2 and here i've got.

03:45 Again this is going to help me or give me my limiting.

03:49 So for my first calculation i'm taking 10 times 8 divided by 58 .1.

03:55 2 divided by 2 and i get 0 .688 moles of co2.

04:11 That's what we could produce with 10 grams of butane and i get 10 .8 10 times 8 divided by 32 divided by 13 i get 0 .192 moles of co2.

04:45 So i can see that my limiting reactant is oxygen, 10 .00 grams, is my limiting reactant.

05:11 That's my limiting reactant.

05:15 And what did i say we needed to figure out? moles of product, theoretical yield for each product.

05:22 So i'm going to, for my first product, i'm going to take 0 .192 moles of co2, and i'm going to multiply this by, hang on, let me go ahead and find i would use 44, but this wants me to use 44 .01.

06:01 So the mass of this reactant is 0 .192 times, and that equals 8 .45 grams of co2.

06:28 I do have more sighings than that here.

06:31 So let me go back and do my times 8...

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