October 2018

# Limiting Reactant

12/10/18 14:30

The answers will be below. Consider the following equation.

C

First you have to balance that equation. Now, imagine I mix 5.45g of C

Answers 1C5H12 + 8O2 --> 5CO2 + 6H2O C5H12 is 72.15 g/mol water is at 18.02g/mol O2 is 32g/mol For C5H12, 5.45g x 1mol/72.15=0.0755mol 11.42g of O2 is 0.3569mol. You clearly have more O2 than C5H12. However, you don't have 8x as much. In other words, for 0.0755 mol of C5H12 you would need 0.604mol. You don't have that, so Oxygen is limiting. So, for 0.3569 mol of oxygen, you would use 1/8 that in C5H12 or 0.0446 mol. To answer the last part first, that means you would have left 0.0755-0.0446 mol of C5H12 which is 0.309 mol or 2.23g left over. For water, you have 0.3569 mol O2 x 6 mol H2O/8molO2 x 18.02g H2O/mol for 4.82g H2O.

C

_{5}H_{12}+ O_{2}--> CO_{2}+ H_{2}OFirst you have to balance that equation. Now, imagine I mix 5.45g of C

_{5}H_{12}and 11.34g of O_{2}How many grams of water will I make? And which ingredient is left over AND how much of that in grams is left?Answers 1C5H12 + 8O2 --> 5CO2 + 6H2O C5H12 is 72.15 g/mol water is at 18.02g/mol O2 is 32g/mol For C5H12, 5.45g x 1mol/72.15=0.0755mol 11.42g of O2 is 0.3569mol. You clearly have more O2 than C5H12. However, you don't have 8x as much. In other words, for 0.0755 mol of C5H12 you would need 0.604mol. You don't have that, so Oxygen is limiting. So, for 0.3569 mol of oxygen, you would use 1/8 that in C5H12 or 0.0446 mol. To answer the last part first, that means you would have left 0.0755-0.0446 mol of C5H12 which is 0.309 mol or 2.23g left over. For water, you have 0.3569 mol O2 x 6 mol H2O/8molO2 x 18.02g H2O/mol for 4.82g H2O.

# Percent Composition

08/10/18 15:35

This contains instructions for determining empirical formula from percent composition. Also there is a stoichiometry problem for practice. Read More…

# The Mole

02/10/18 14:12

Introduction to the concept of the Mole (mol). It's just a unit of stuff…like a dozen, but bigger. Read More…