Ammonia and oxygen react to form nitrogen and water. 4 NH3(g)+3 O2(g) ⟶2 N2(g)+6 H2 O(g) a. How

step 1 Okay, so for this problem, there are three different parts. It's going to be three different str

Ammonia and oxygen react to form nitrogen and water. 4...

Ammonia and oxygen react to form nitrogen and water. $$4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ a. How many grams of $\mathrm{O}_{2}$ are needed to react with $13.6 \mathrm{~g}$ of $\mathrm{NH}_{3}$ ? b. How many grams of $\mathrm{N}_{2}$ can be produced when $6.50 \mathrm{~g}$ of $\mathrm{O}_{2}$ reacts? c. How many grams of $\mathrm{H}_{2} \mathrm{O}$ are formed from the reaction of $34.0 \mathrm{~g}$ of $\mathrm{NH}_{3}$ ?

Ammonia and oxygen react to form nitrogen and water.
$$4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$
a. How many grams of $\mathrm{O}_{2}$ are needed to react with $13.6 \mathrm{~g}$ of $\mathrm{NH}_{3}$ ?
b. How many grams of $\mathrm{N}_{2}$ can be produced when $6.50 \mathrm{~g}$ of $\mathrm{O}_{2}$ reacts?
c. How many grams of $\mathrm{H}_{2} \mathrm{O}$ are formed from the reaction of $34.0 \mathrm{~g}$ of $\mathrm{NH}_{3}$ ?

Key Concepts

Stoichiometric Ratios

Derived from the balanced chemical equation, stoichiometric ratios are used to relate the number of moles of one substance to the number of moles of another substance in the reaction. They allow for the determination of how much of one reactant is required to completely react with another or how much product can be formed.

Conversion between Grams and Moles

This process involves using the molar mass to convert a given mass of a substance into moles, or vice versa. It is a fundamental calculation in chemistry that is applied to determine the amount of reactants or products involved in a reaction.

Dimensional Analysis

Dimensional analysis is a systematic approach to unit conversion that helps ensure consistency and correctness in stoichiometric calculations. By canceling units appropriately, it bridges the gap between different unit systems, such as grams and moles, in chemical calculations.

Balanced Chemical Equation

A balanced chemical equation is essential for stoichiometry, as it provides the exact mole ratios of reactants and products. This ensures that the law of conservation of mass is met and allows for accurate conversion between amounts of substances present in a reaction.

Molar Mass

The molar mass of a compound, expressed in grams per mole, is a key concept in converting between the mass of a substance and the number of moles. It is derived from the atomic masses of the elements in the compound and is essential for quantitative calculations in chemical reactions.

Transcript

00:01 Okay, so for this problem, there are three different parts.

00:04 It's going to be three different strychialmetry problems.

00:07 So for part a, we are starting with 13 .6 grams of nh3, and it's asking us to solve to grams of 02.

00:23 So friendly reminder, grams to grams has to be three steps.

00:28 We have to first convert to moles, then use our molar ratio.

00:30 And finally, we have to convert back to grams.

00:35 So when we do this, the first thing we're going to do, we start with grams of nh3.

00:42 And we already know there's going to be one, two, three steps.

00:54 So the first step, we have to get rid of this gram.

00:56 So grams of nh3, we go on the bottom.

01:00 And the only thing that can go with grams of nh3 is one mole of nh3.

01:04 So one mole of nh3 goes on top to find the molar mass.

01:08 Use the periodic table.

01:10 And if you use the periodic table, you find the mass to be 17 .04.

01:16 To indicate that this is coming from something, let me get a different color.

01:21 So this is 17 .04.

01:25 So this is periodic table.

01:28 So if it's in blue, it's from the periodic table.

01:30 Okay.

01:31 So we've gotten rid of grams.

01:33 We're now at moles.

01:34 So now we're going to use moles in the bottom of this second fraction.

01:38 So this is going to be our molar ratio.

01:40 So we're going to put moles of nh3 on the bottom because that's what we have to get rid of.

01:43 We're trying to get to grams of o2, but we can't go straight to grams yet.

01:47 Instead, we're going to go to moles of o2.

01:51 So when we see mole on top of mole, we have to go up to our balanced chemical reaction.

01:58 So if we look, o2 has a three out front, nh3 has a four out front.

02:06 Therefore, it's going to be a three with o2, and it's going to be a four, with nh3.

02:14 And then last but not least, we need to convert this moles of 02.

02:17 So it's going to be one mole of 02 because we're trying to get grams of 02.

02:23 So this is again where we would use our periodic table.

02:26 And 16 times 2 gives us a mass of 32.

02:30 So then we're going to multiply straight across the top, multiply straight across the bottom, divide those two numbers.

02:36 And when we divide them, we get 19 .2 grams of 02 .2.

02:42 So that is the answer for.

02:44 Part a.

02:47 For part b, it's asking us to determine the amount of n2 produced from 6 .5 grams of not n, i'm so sorry, from 02, so 6 .5 grams of 0 to grams of n2 produced.

03:13 Again, with grams to grams, it's going to be three steps just like it was above.

03:17 So the first step is going to be converting this grams to moles...

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