How To Find The Amount of Excess Reactant That Is Left Over - Chemistry

Example 2: Calculating Excess Reactant Consumed00:10:33

Starting with 70 grams of sulfur (the limiting reactant), a gram-to-gram conversion is performed to find the amount of aluminum consumed. It is calculated that 39.26 grams of aluminum are consumed.

Example 2: Determining the Remaining Excess Reactant00:12:25

From the initial 50 grams of aluminum, 39.26 grams are consumed, leaving 10.74 grams of aluminum as the excess reactant.

Introduction to Finding Excess Reactant00:00:01

The video introduces the objective: to find the mass of the excess reactant remaining after a reaction. The process is broken down into three steps: identifying the limiting reactant, determining the mass of the excess reactant consumed, and calculating the final remaining mass.

Example 1: Balancing the Chemical Equation (Aluminum and Sulfuric Acid)00:00:39

The first example involves aluminum reacting with sulfuric acid to produce hydrogen gas and aluminum sulfate. The chemical formula for aluminum sulfate (Al2(SO4)3) is derived, and the equation is balanced.

Example 1: Identifying the Limiting Reactant00:01:57

Given 80 grams of aluminum and 40 grams of sulfuric acid, the limiting reactant is identified. The masses are converted to moles (2.965 moles for Al, 0.407 moles for H2SO4), and then divided by their stoichiometric coefficients to determine that sulfuric acid is the limiting reactant.

Example 1: Calculating Excess Reactant Consumed00:04:16

Using the amount of the limiting reactant (40g H2SO4), a gram-to-gram conversion is performed to find the amount of aluminum consumed. It is calculated that 7.33 grams of aluminum are consumed in the reaction.

Example 1: Determining the Remaining Excess Reactant00:06:03

With 80 grams of aluminum initially and 7.33 grams consumed, the remaining mass of aluminum is calculated as 72.67 grams.

Example 2: Balancing the Chemical Equation (Aluminum and Sulfur)00:07:06

A second practice problem is introduced: 50 grams of aluminum reacting with 70 grams of elemental sulfur (S8) to produce aluminum sulfide. The chemical formula for aluminum sulfide (Al2S3) is determined, and the equation is balanced.

Example 2: Identifying the Limiting Reactant00:08:19

The masses (50g Al, 70g S8) are converted to moles (1.853 moles for Al, 0.2728 moles for S8). Dividing by their coefficients, it's determined that sulfur is the limiting reactant, despite starting with a greater mass of sulfur.