One of the steps in the commercial process for converting...

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO as shown by the balanced chemical equation below: 4 NH3(g) + 5 O2(g) → 4 NO (g) + 6 H2O(g) In a certain experiment, 25.0 g of NH3 is allowed to react with 65.0 g of O2. A What is the limiting reactant? answer choices- (O2, NH3, NO, H2O) B the mass of NO produced is? (in g) answer choices -(44.0, 1.47, 48.8, 2.03) C is the excess reactant? answer choices- (O2, NH3, NO, H2O) D grams of it remains at the end of the reaction? answer choices (58.7, 40.0, 6.31, 21.0)

Transcript

00:02 Given the balanced chemical equation, find the limiting reactant, the mass of one of the products, the excess reactant, and how many grams of the excess reactant that remain.

00:15 So our balanced chemical equation is we have four ammonia, and that's going to be in gas, and that's reacting with five oxygens, also in gaseous form, and that is forming 4 .mone.

00:35 Plus six waters, also in a gas form.

00:44 We're told we have 25 .0 grams of ammonia reacting with 65 .0 grams of oxygen.

00:56 We need to find the number of grams of no that will be made along with the limiting reactant.

01:05 So let's convert our grams to moles.

01:08 So we have 25 .0 grams of ammonia.

01:16 Ammonia weighs 17 .01 grams for every one mole.

01:27 Now let's convert this into moles of no.

01:31 So we know for every four moles of ammonia that we have, we will get four moles of n0.

01:42 So now we'll be in moles of no.

01:44 So let's cancel everything out.

01:47 Let's do the same thing with oxygen.

01:49 We have 65 .0 grams and one mole of oxygen weighs 32 .0 grams.

02:00 Now our stoichiometric ratio is going to be a little bit different.

02:04 We're going to have five moles of oxygen on the bottom and four moles of no on the top.

02:11 So moles of oxygen will cancel and grams of oxygen will cancel.

02:16 So let's go back up to ammonia for a second.

02:19 We have 25 .0 divided by 17 .031 and then multiplied by 4 over 4, or you can reduce that down to 1 over 1.

02:32 And you're going to get 1 .4679 moles of n0.

02:39 Now, if we consider oxygen, we have 65 .0 divided by 32 times 4 over 5, that's going to give us 1 .625 moles of no...

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