### Video Transcript

According to the given equation, how many moles of HCl are required to react completely with 1.5 moles of NF₃? A) 0.50 moles, B) 1.0 moles, C) 1.5 moles, D) 2.0 moles, E) 4.5 moles.

We have an unbalanced equation here. So our priority is to balance the equation first, using balancing numbers to find the mole ratios or stoichiometry for this equation. Stoichiometric coefficients or balancing numbers in a balanced equation are usually whole numbers that indicate the mole ratios in the equation. These are the proportions in which the species react in a given reaction. Notice in the question here, they’re missing. A quick inspection of how many atoms of each type of element there are present on each side of this equation quickly shows us that this equation is not balanced.

There was an imbalance in the amount of nitrogen, fluorine, and chlorine atoms on each side of this equation, as it stands at the moment. The easiest way to balance this equation is to deal with nitrogen first, as it appears as a diatomic element on the right-hand side. We need two atoms of nitrogen on the left-hand side. So we use a coefficient of two just before the NF₃. This step also introduces six fluorine atoms on the left-hand side. So we need to use a balancing coefficient of six just before the HF molecule on the right-hand side of the equation. We’ve now balanced the nitrogen atoms and the fluorine atoms in this equation. But we need to check the other atoms before we’re satisfied that this equation is actually balanced.

We now have six hydrogen atoms on the right-hand side of the equation. So we need to place a six before the HCl to balance the hydrogen on the left hand-side of the equation. Finally, as chlorine appears as a diatomic element on the right-hand side of the equation, it’s easiest to balance this last of all. We need three molecules of Cl₂ to balance the six atoms of chlorine introduced on the left-hand side. This equation is now balanced. From our balanced equation, we can see that two moles of NF₃ reacts completely with six moles of HCl. The ratio of two to six can be simplified as one to three by dividing both sides by two. We now know that one mole of NF₃ requires three moles of HCl for a complete reaction.

In the question, we’re being asked about what happens when 1.5 moles of NF₃ reacts with HCl. We need to scale our ratio up by a factor of 1.5 on both sides. By using the scaling factor of 1.5 on both sides of our ratio, we discover that 4.5 moles of HCl are required. 4.5 moles of HCl is therefore the correct answer.