Write a balanced chemical equation for each chemical...

Write a balanced chemical equation for each chemical reaction. (a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide. (b) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas. (c) Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas. (d) Liquid benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$ reacts with gaseous oxygen to form carbon dioxide and liquid water.

Write a balanced chemical equation for each chemical reaction.
(a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide.
(b) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.
(c) Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas.
(d) Liquid benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$ reacts with gaseous oxygen to form carbon dioxide and liquid water.

Key Concepts

Chemical Reaction Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. A balanced chemical equation serves as the foundation for stoichiometric calculations, providing the mole ratios needed to determine the amounts of substances consumed and produced.

Phase Indicators in Chemical Equations

Phase indicators, such as (s) for solids, (l) for liquids, (g) for gases, and (aq) for aqueous solutions, specify the physical states of the substances in a chemical reaction. Including these notations provides additional context about the reaction conditions and helps in accurately representing the reaction process.

Balancing Chemical Equations

Balancing chemical equations involves assigning coefficients to reactants and products so that the number of atoms for each element is equal on both sides. This ensures that the reaction adheres to the law of conservation of mass by accurately representing the amounts of each substance involved.

Law of Conservation of Mass

The law of conservation of mass states that matter cannot be created or destroyed during a chemical reaction. This principle underpins the need to balance equations, as every atom present in the reactants must be present in the products.

Transcript

00:01 So in the first case we have given that solid copper reacts with solid sulfur to form copper first sulfide so that can be written as cu2s okay and that is again solid now we need to balance this equation so here on the right side i have two copper atoms okay so here we also need to okay so now we have equal number of copper atoms on both sides and we have equal number of sulfur sulfur atoms on both sides so the equation is balanced.

00:33 Ok so you have to make sure that the number of atoms on each side of reactant and product are equal ok.

00:40 Now for part 2 this is a and b here actually.

00:48 So in b part we have sulfur dioxide gas that reacts with oxygen gas to form sulfur trioxide side gas okay so now we need to balance this equation so we have one sulfur atom on each side of reactant and product so they are equal we have two oxygen atom plus one of here two oxygen atom so that is total of four oxygen atoms on left side but three oxygen atom only on the right side so, we need to balance this.

01:30 If i multiply this by 2, so then we have 6 oxygen atom.

01:35 If i multiply this by 2, then we have 4 oxygen atom plus 2 oxygen atom, so 6 oxygen atom.

01:42 So, now the oxygen atoms are equal also.

01:46 So, this equation is balanced.

01:49 But wait, we have now 2 sulfur atoms so i need to balance this also okay so now for this i have to multiply this by 2 now by doing this i have 4 oxygen atom plus 4 oxygen atom that is 8 oxygen atom but in this case i have only 6 oxygen atom okay so now removing the coefficient from here now you see that i have 4 oxygen atom plus 2 oxygen atom so 6 oxygen atom and this is 3 multiplied by 2 6 oxygen atom so now this equation is balanced okay moving on to c part so we have hydrochloric acid in a coarse form that is hcl okay and it reacts with manganese oxide mn4o that is in solid form and this gives mncl2 in aqueous form plus liquid water h2o in liquid state plus cl2 in gaseous form.

03:17 Form.

03:18 Okay.

03:25 So now if i multiply this by 2 so i have 2 hydrogen atom on left side also and 2 hydrogen atom on right side also.

03:35 Also for chlorine atoms we have 2 chlorine atoms on left side and 2 chlorine atoms on right side but here i have 2 chlorine atoms in this case.

03:44 Okay.

03:44 So that is total of 4 chlorine atoms on right side are present...

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