Question

1. to balance a chemical equation, it may be necessary to adjust the

a. coefficients
b. subscripts
c. superscripts
d. formulas of the compounds
part 2 balancing and classifying: balance and classify the following chemical reactions. write your classification in the blank to the right. (4 pts each, unless otherwise indicated, 41 pts total)

1. __licl + br₂ → libr + __cl₂

1. __kclo₃ → kcl + __o₂

1. __n₂ + o₂ → __n₂o₅

1. __c₃h₈ + o₂ → co₂ + __h₂o

1. __ca(oh)₂ + h₃po₄ → ca₃(po₄)₂ + __h₂o

1. write the chemical equation for the following reaction. then balance and classify. (9 pts)

aqueous potassium hydroxide reacts with aqueous copper (ii) sulfate to make aqueous copper (ii) hydroxide and solid potassium sulfate.

Explanation:

Question 25

To balance a chemical equation, we adjust the coefficients (the numbers in front of compounds) to ensure the same number of each atom on both sides. Subscripts represent the ratio of atoms in a compound and can't be changed (as that would change the compound itself), superscripts are for charges, and we don't change compound formulas. So the correct answer is the one about adjusting coefficients.

Step 1: Balance Li and Cl

On the left, we have 1 Li and 1 Cl (from LiCl). On the right, 1 Li (LiBr) and 2 Cl (Cl₂). So we need 2 LiCl to get 2 Cl. Now the equation is \( 2\text{LiCl} + \text{Br}_2
ightarrow \text{LiBr} + \text{Cl}_2 \).

Step 2: Balance Li and Br

Now we have 2 Li on the left, so we need 2 LiBr on the right. Then Br: left has 2 (Br₂), right has 2 (2 LiBr). Now the balanced equation is \( 2\text{LiCl} + \text{Br}_2
ightarrow 2\text{LiBr} + \text{Cl}_2 \).

Step 3: Classify the reaction

This is a single - replacement reaction (one element replaces another in a compound: Cl⁻ in LiCl is replaced by Br₂, or Br₂ replaces Cl⁻? Wait, actually, LiCl + Br₂ → LiBr + Cl₂. Here, Cl in LiCl is replaced by Br? No, wait, Br₂ is a diatomic molecule. Actually, the more accurate way: in single - replacement, a more reactive element replaces a less reactive one. The general form is \( A + BC
ightarrow AC + B \) (if A is a metal) or \( A + BC
ightarrow BA + C \) (if A is a non - metal). Here, Cl (in LiCl) is replaced by Br? Wait, no, the oxidation states: Li is +1, Cl is - 1, Br is 0 in Br₂, +1 in LiBr, 0 in Cl₂. So Cl⁻ is oxidized to Cl₂ (losing electrons), Br₂ is reduced to Br⁻ (gaining electrons). Wait, maybe I got the direction wrong. Let's see: LiCl (aq) + Br₂ (l) → LiBr (aq) + Cl₂ (g). But in reality, Cl is more reactive than Br (halogen reactivity: F>Cl>Br>I), so this reaction as written doesn't occur. But assuming we are just balancing, the type is single - replacement (even though thermodynamically it's not favorable, the form is single - replacement: non - metal replaces non - metal in a compound).

Step 1: Balance K and Cl

Left: 1 K, 1 Cl (KClO₃). Right: 1 K, 1 Cl (KCl). So K and Cl are balanced for now.

Step 2: Balance O

Left: 3 O (KClO₃). Right: 2 O (O₂). The least common multiple of 3 and 2 is 6. So we need 2 KClO₃ (to get 6 O) and 3 O₂ (to get 6 O). Now the equation becomes \( 2\text{KClO}_3
ightarrow \text{KCl}+ 3\text{O}_2 \).

Step 3: Balance K and Cl

Now we have 2 K on the left (2 KClO₃), so we need 2 KCl on the right. The balanced equation is \( 2\text{KClO}_3
ightarrow 2\text{KCl}+ 3\text{O}_2 \).

Step 4: Classify the reaction

This is a decomposition reaction (a compound breaks down into simpler substances: \( \text{KClO}_3 \) breaks down into KCl and O₂).

Answer:

A. coefficients

Question 26