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Question

2 C₄H₁₀ + 13 O2 → 8 CO₂ + 10 H₂O

How many grams of C₄H₁₀ are needed to react with 51.40 grams of O₂? 

Answer 1

First let's start by converting the 51.40 grams of O₂ into moles of O₂. For this conversion, we will need to find the molar mass of O₂. The molar mass is the number of grams of O₂ in one mole of O₂. To find the molar mass, let's go to the periodic table and find the atomic mass of O (oxygen). The atomic mass of oxygen is 15.999. So one O = 15.999. But if you look at O₂, you will notice the subscript after the O. The subscript is 2. So this means we have 2 Os. If one O = 15.999, then 2 Os = 2 x 15.999. The calculator says that

2 x 15.999 = 31.998. This is the molar mass of O₂. In other words, 1 mole of O₂ = 31.998 grams of O₂. So now we need to convert our 51.40 grams of O₂ into moles of O₂. We set up the conversion like this:

51.40 grams of O₂ x 1 mole of O₂

31.998 grams of O₂

Notice that we have "grams of O₂" on the top and the bottom. This allows the "grams of O₂" to cancel out, so we will be left with moles of O₂, which is exactly what we want.

Now we multiply straight across. In other words, we do 51.40 x 1. Obviously, 51.40 x 1 = 51.40. Next we divide 51.40 by 31.998. The calculator says that 51.40 / 31.998 = 1.60635. So we have 1.60635 moles of O₂. (In other words, 51.40 grams of O₂ = 1.60635 moles of O₂).

Now that we have converted the number we were given (51.40 grams of O2) into moles, let's look at the balanced equation the problem gave us:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O

Remember that the numbers in front of the compounds or elements are called coefficients. For example, the coefficient of C₄H₁₀ is 2. The coefficient of O₂ is 13. These two coefficients are going to be very important in the next step, so remember them. Why these two coefficients in particular? Because we were given grams of O₂ in the problem, and the problem asks us to find grams of C₄H₁₀. So we will need to know the coefficients of O₂ and C₄H₁₀ to solve the problem.

The coefficients tell us how many moles of one element or compound are needed to create a certain number of moles of another compound. For example, it takes 2 moles of C₄H₁₀ to create 13 moles of O₂. It takes 2 moles of C₄H₁₀ to create 8 moles of CO₂. But we only focus on the C₄H₁₀ and O₂ in this problem.

Now let's set up our formula to convert moles of O₂ into moles of C₄H₁₀. The formula looks like this:

moles of O₂ x coefficient of C₄H₁₀

coefficient of O₂

We already know the moles of O₂ because we calculated it already. Remember that we have 1.60635 moles of O₂. We also already found the coefficient of C₄H₁₀ in the balanced equation. The coefficient is 2. And the coefficient of O₂ in the balanced equation is 13. We put this information into the formula like this:

1.60635 moles of O₂ x 2

13

So now we use a calculator to calculate this formula. We will be doing 1.60635 x (2/13). The calculator says that 1.60635 x (2/13) = 0.24713. This number is the number of moles of C₄H₁₀ needed to react with 51.40 grams of O₂. So, 0.24713 moles of C₄H₁₀ are needed to react with 51.40 grams of O₂.

Unfortunately, the problem asked for grams of C₄H_10. So we need to convert 0.24713 moles of C₄H₁₀ into grams of C4H10. For this conversion, we need to find the molar mass of C₄H₁₀. We start by looking up the atomic mass of C (carbon) in the periodic table. The atomic mass of one C is 12.01. But the subscript for C in C₄H₁₀ is 4, so we have 4 Cs. We need to find the mass of all 4 Cs. So if the atomic mass of one C is 12.01, then the mass of 4 Cs is 4 x 12.01. The calculator says that 4 x 12.01 = 48.04.

Now we move on to the next element in C₄H₁₀, which is H (hydrogen). We look up the atomic mass of H in the periodic table. The atomic mass of one H is 1.008. But the subscript for H in C₄H₁₀ is 10, so we have 10 Hs. We need to find the mass of all 10 Hs. So if the atomic mass of one H is 1.008, then the mass of 10 Hs is 10 x 1.008. The calculator says that 10 x 1.008 = 10.08.

So far we have found the mass of the four Cs and the 10 Hs in C₄H₁₀. To find the molar mass of C₄H₁₀ as a whole, we simply add mass of the four Cs + the mass of the 10 Hs. The mass of the 4 Cs is 48.04 and the mass of the 10 Hs is 10.08. So we do 48.04 + 10.08. The calculator says that 48.04 + 10.08 = 58.12. So, the molar mass of C₄H₁₀ is 58.12. In other words, there are 58.12 grams of C₄H₁₀ in one mole of C₄H₁₀.

Now we convert the 0.24713 moles of C₄H₁₀ to grams of C₄H₁₀. We set up our formula like this:

0.24713 moles of C₄H₁₀ x 58.12 grams of C₄H₁₀

1 mole of C₄H₁₀

To solve this, we just multiply 0.24714 x 58.12. Notice that "mole(s) of C₄H₁₀" is on the top and bottom of this formula. This allows the "mole(s) of C₄H₁₀" to cancel, and we are left with "grams of C₄H₁₀", which is exactly what we want.

Now let's multiply 0.24714 x 58.12. The calculator says that 0.24714 x 58.12 = 14.36. In other words, 14.36 grams of C₄H₁₀ are needed to react with 51.40 grams of O₂.

Final answer: 14.36 grams of C₄H₁₀ are needed