When 12.8 g of Na and 10.2 g of Cl2 react, what is the mass, in grams, of NaCl that is produce

When 12.8 g of Na and 10.2 g of Cl2 react, what is the...

Key Concepts

Balanced Chemical Equation

A balanced chemical equation shows the proportions of reactants and products involved in a chemical reaction. It is essential to ensure that the law of conservation of mass is followed, meaning the number of atoms of each element is the same on both sides of the equation.

Molar Mass Calculation

Molar mass calculation involves determining the mass of one mole of a substance, usually expressed in grams per mole. This concept is crucial for converting between the mass of a substance and the number of moles, which is fundamental in quantitative chemical analysis.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction as defined by the balanced chemical equation. It enables the calculation of the amounts of reactants required or products formed in a reaction, based on the mole ratios derived from the equation.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that runs out first, thereby limiting the amount of product that can be formed. Determining the limiting reactant is essential for calculating the theoretical yield of the reaction.

Transcript

00:01 Okay, so in this problem we are trying to figure out the mass of n, a, b, r in grams.

00:06 So we can always set up a system of equations by adding in the mass of one mixture and adding in the mass of a second mixture, and we know that's going to come out to the total mass of the mixture combined.

00:24 And in this case we know that is going to be 2 grams.

00:29 2 .00 grams.

00:31 So in the case of this problem i'm going to let m1 represent the mass of n, a, c, l, and i'm going to have m2 be what we're trying to find out, the mass of n, a, b, r.

00:52 Now one other thing we know about this problem is that this mixture is going to contain 0 .75 grams of sodium, and we know that sodium or the n, a is going to come from both n, a, c, l and n, a, b, r.

01:07 So we need to set up an equation for the n, a or sodium content within the whole mixture.

01:14 So to do that we are going to say 0 .75 grams of sodium is equal to the mass fraction of our sodium in n, a, c, l.

01:36 We're going to multiply that by the mass of mixture one, and that is going to be added to the mass fraction of sodium in n, a, b, r.

01:52 And since that we labeled as m2, we're going to multiply that by m2.

01:58 So we know that the mass fraction of sodium in n, a, c, l is going to be the molar mass of sodium divided by the molar mass of n, a, c, l.

02:08 And we're going to do the same thing for n, a, b, r.

02:12 So we have 0 .75 grams sodium is equal to, and i'm going to use brackets to make this more clear, 23 grams per mole divided by 58 .5 grams per mole, and that whole thing is going to be multiplied by m1.

02:40 Then we're going to again do the same thing for n, a, b, r.

02:45 We have 23 grams per mole divided by 102, parenthesis, 102 .9 grams per mole, and that is going to be multiplied by m2.

03:04 So we can actually use this equation up here to find a substitute for m1 in this equation that we just created.

03:14 So if we manipulate this equation up here a little bit, we can say that m1 is equal to 2 grams minus m2.

03:28 So we are going to fill this in for m1 in this equation we just created down here...

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