How many grams of each product result from the following...

How many grams of each product result from the following reactions, and how many grams of which reactant is left over? (a) $(1.3 \mathrm{g} \mathrm{NaCl})+\left(3.5 \mathrm{g} \mathrm{AgNO}_{3}\right) \longrightarrow(x \mathrm{g} \mathrm{AgCl})+\left(y \mathrm{g} \mathrm{NaNO}_{3}\right)$ (b) $\left(2.65 \mathrm{g} \mathrm{BaCl}_{2}\right)+\left(6.78 \mathrm{g} \mathrm{H}_{2} \mathrm{SO}_{4}\right) \longrightarrow\left(x \mathrm{g} \mathrm{BaSO}_{4}\right)+(y \mathrm{g} \mathrm{HCl})$

How many grams of each product result from the following reactions, and how many grams of which reactant is left over?
(a) $(1.3 \mathrm{g} \mathrm{NaCl})+\left(3.5 \mathrm{g} \mathrm{AgNO}_{3}\right) \longrightarrow(x \mathrm{g} \mathrm{AgCl})+\left(y \mathrm{g} \mathrm{NaNO}_{3}\right)$
(b) $\left(2.65 \mathrm{g} \mathrm{BaCl}_{2}\right)+\left(6.78 \mathrm{g} \mathrm{H}_{2} \mathrm{SO}_{4}\right) \longrightarrow\left(x \mathrm{g} \mathrm{BaSO}_{4}\right)+(y \mathrm{g} \mathrm{HCl})$

Key Concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It involves using a balanced chemical equation to determine the relative number of moles of each substance involved, which can then be converted into masses using the appropriate molar masses.

Limiting Reagent

The limiting reagent is the reactant that is completely consumed first during a chemical reaction, thereby limiting the amount of product that can be formed. Identifying the limiting reagent is essential in determining the theoretical yield of products and in calculating any remaining excess reactant.

Molar Mass

Molar mass is the mass of one mole of a given substance, calculated by summing the atomic masses of the constituent elements. This conversion factor is crucial in stoichiometric calculations to convert between the mass of a substance and the number of moles, allowing the use of mole ratios from the balanced equation.

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction with equal numbers of atoms for each element on both the reactant and product sides, ensuring the law of conservation of mass is obeyed. This equation provides the mole ratios needed for stoichiometric calculations to determine how much product is theoretically formed from given quantities of reactants.

Excess Reactant Calculation

Excess reactant calculation involves determining the amount of reactant that remains unreacted after the completion of the reaction. Once the limiting reagent has been identified and the moles of product formed are calculated, the consumption of the excess reagent can be subtracted from the original amount to find how much is left over.

Transcript

00:03 To answer this question, we need to start with mole ratios.

00:07 To do that, we're going to first divide each of the grams of the reactants by their respective molar masses.

00:14 The molar mass of sodium chloride nacl is 58 .44 grams for every one mole, which means that we have 0 .022 -2 moles of nacl.

00:32 We're going to do similar for silver nitrate, dividing by its molar mass, which is 169 .87.

00:46 That's equal to 0 .026 moles of silver nitrate.

00:54 Take a look at the reaction and realize that there are secret coefficients of one on each to make a balanced reaction.

01:04 That means that you need equal mole ratios, one to one more ratio.

01:09 Of these react.

01:11 So if we compare the number of moles of sodium chloride and silver nitrate, we see that there is less silver nitrate, and that makes silver nitrate the limiting reactants.

01:27 So this is the limiting reactants.

01:29 It will be used up completely in the reaction, and we're going to have some left over sodium chloride.

01:38 So 0 .026 moles of silver nitrate will react with 0 .02 ,000 ,000 ,000 ,000 ,000 ,000, 0 .26 -206 moles of sodium chloride.

01:50 So this is how much is available.

01:57 The 0 .206 is how much sodium chloride is used.

02:02 And if we subtract these, we get 0 .016 moles of sodium chloride left over.

02:14 To find the mass of sodium chloride that's left over, we're going to multiply this number of moles by the molar mass, which we already found.

02:24 Grants for every mole.

02:27 And see that we have 0 .96 grams of sodium chloride left over.

02:36 First part of the problem.

02:38 Next, moving into finding grams of each product.

02:42 We're going to use this number of moles, which leads us to identify that if this many moles of the silver nitrate are used, we're going to have as many moles of silver chloride and the same number of moles of sodium nitrate produced.

03:13 We'll do this a little differently this way...

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