For each of the following balanced equations, calculate how...

For each of the following balanced equations, calculate how many grams of each product would be produced by complete reaction of 15.0 $\mathrm{g}$ of the reactant indicated in boldface. a. $2 \mathbf{B C l}_{3}(s)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{B}(s)+6 \mathrm{HCl}(g)$ b. $2 \mathbf{C u}_{2} \mathbf{S}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}(s)+2 \mathrm{SO}_{2}(g)$ c. $2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathbf{C} \mathbf{u}_{2} \mathbf{S}(s) \rightarrow 6 \mathrm{Cu}(s)+\mathrm{SO}_{2}(g)$ d. $\mathrm{CaCO}_{3}(s)+\mathbf{S} \mathbf{i} \mathbf{O}_{2}(s) \rightarrow \mathrm{CaSiO}_{3}(s)+\mathrm{CO}_{2}(g)$

For each of the following balanced equations, calculate how many grams of each product would be produced by complete reaction of 15.0 $\mathrm{g}$ of the reactant indicated in boldface.
a. $2 \mathbf{B C l}_{3}(s)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{B}(s)+6 \mathrm{HCl}(g)$
b. $2 \mathbf{C u}_{2} \mathbf{S}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}(s)+2 \mathrm{SO}_{2}(g)$
c. $2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathbf{C} \mathbf{u}_{2} \mathbf{S}(s) \rightarrow 6 \mathrm{Cu}(s)+\mathrm{SO}_{2}(g)$
d. $\mathrm{CaCO}_{3}(s)+\mathbf{S} \mathbf{i} \mathbf{O}_{2}(s) \rightarrow \mathrm{CaSiO}_{3}(s)+\mathrm{CO}_{2}(g)$

Key Concepts

Balanced Chemical Equations

Balanced chemical equations represent the conservation of mass and atoms by showing the same number of each atom on both sides of the equation. They provide the stoichiometric coefficients that indicate the proportional relationships between the amounts of reactants and products, which are essential for any quantitative chemical calculation.

Mole Concept

The mole concept is a fundamental idea in chemistry that relates the number of particles in a substance to its mass. It serves as the bridge between the atomic scale and the laboratory scale, enabling chemists to convert between the number of entities (atoms, molecules, ions) and measurable mass.

Molar Mass

Molar mass is the mass of one mole of a chemical compound, typically expressed in grams per mole (g/mol). It allows for the conversion between the mass of a substance and the number of moles, which is crucial for relating the amounts of reactants and products in chemical reactions.

Stoichiometric Calculations

Stoichiometric calculations involve using the mole ratios from a balanced chemical equation to determine the quantitative relationships between reactants and products. By converting masses to moles, applying the mole ratios, and then converting back to grams, one can calculate the expected amounts of products formed or reactants required in a chemical reaction.

Transcript

00:01 Hi everyone, you know in first case that is 2 bcl 3 gives a 2 boron and 6hcl and that's why now 2 into molecular mass of a boron trichloride that is 117 .17 gram of bcl3 b .c .l .3 corresponds to 2 into 10 .811 gram of boron and therefore 15 gram of basal 3 corresponds to you can say x gram of boron.

01:01 And therefore amount of boron produced, that is equal to 2 into 10 .811 into 15.

01:17 Divided by 2 again divided by 117 .17 nothing but 1 .384 gram this is what the answer of first part secondly you know 2 into 117 .17 gram of boron trichloride corresponds to 6 into 36 .5 gram of scl therefore 15 gram of boron trichloride corresponds to you can say x gram of hcl and therefore amount of hcl produced from 15 gram of bcl 3 that is equal to 14 .02 gram.

03:07 Then you know b, you know, two moles of co2s gives two moles of co2 and two modes of so2.

03:19 So, first part, that is 2 into molecular mass of co2 s is 159 .157 gram of co2 corresponds to 2 into molecular mass of co2o 143 .0 914 gram of co2 and therefore 15 gram of cos corresponds to x gram of cos and therefore amount of ce2s produced amount of c2s to s produced that is equal to 13 .149 gram.

04:42 This is the answer.

04:44 Then we will go for the second that is 2 into 159.

04:56 157 gram of co2s corresponds to 2 into 64 gram of so2.

05:07 64 gram of so2.

05:12 And therefore 15 gram of co2s corresponds to x gram of s and therefore amount of so2 produced that is equal to 6 .0332 gram.

05:59 The next one is 2 into 159 .157 gram of co2 s.

06:20 Gives 6 mol of copper and 1 mole of s .o2...

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