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Video instructions and help with filling out and completing Who Form 5495 Limitation

Instructions and Help about Who Form 5495 Limitation

In this video, we're going to work through a practice problem on the limiting reactant and excess reactant. This problem is pretty straightforward, so if you're just starting out learning limiting reactant, this is a great place to begin. We're going to use this equation to answer the following questions: 1. What is the greatest amount of MgO (magnesium oxide) in moles that can be made with 7.8 moles of Mg (magnesium) and 4.7 moles of O2 (oxygen)? 2. Which is the limiting reactant and which reactant is in excess? 3. How many moles of the excess reactant are left over? This problem is all about moles - how many moles of each reactant we have and how many moles of the product we can make. We have 7.8 moles of magnesium and 4.7 moles of oxygen. To find the greatest amount of magnesium oxide we can make, we need to figure out which reactant is the limiting reactant - is it magnesium or oxygen? To determine the limiting reactant, we'll ask a few questions. First, we'll ask how many moles of oxygen we need if we use all of the magnesium. Then, we'll ask how many moles of magnesium we need if we use all of the oxygen. We'll compare the answers to these questions to determine which reactant runs out first. Let's start with the first question: To use all of the magnesium, how many moles of oxygen do we need? We start with 7.8 moles of magnesium and use the conversion factor from the equation: 2 moles of magnesium for every 1 mole of oxygen. This gives us: (7.8 moles Mg) x (1 mole O2 / 2 moles Mg) = 3.9 moles O2 So, we need 3.9 moles of oxygen to use all of the magnesium. Now, let's ask the reverse question: To use all of the...