Question

1. when balancing a chemical reaction, what are the small numbers that we do not change?
2. according to the law of conservation of mass, can atoms be destroyed during a reaction? explain.

in the questions below, put a checkmark next to the equations that violate the law are put balanced correctly. check each element.

1. ____ 2h₂ + o₂ → 2h₂o
2. ____ na + o₂ → na₂o
3. ____ kcl + br₂ → kbr + cl₂
4. ____ 2 h₂ + o₂ → 2 h₂o
5. ____ na₂co₃ + 2hcl → 2nacl + h₂o + co₂
6. ____ 4 h₂ + 2 o₂ → 4 h₂o
7. ____ 2fe₂o₃ + 3c → 4fe + 3co₂
8. ____ 2h₂ + o₂ → h₂o

use the following reaction to answer the next questions.
h₂o(aq) → h₂o(l) + o₂(g)

1. in the reaction above, what does the (l) stand for next to h₂o?
2. in the reaction above, what does the (g) stand for next to h₂o?
3. in the reaction above, what does the (aq) stand for next to h₂o?
4. knowing this information above, what would an (s) symbol stand for?
5. in the reaction above, list the reactants):
6. in the reaction above, list the products):

Explanation:

Step1: Analyze the reaction

The given reaction is \( \ce{H_{2}O(aq)
ightarrow H_{2}O(l) + O_{2}(g)} \). Reactants are the substances on the left - hand side of the reaction arrow, and products are on the right - hand side.

Step2: Identify reactants

For the reactant, we look at the left side of the arrow. So the reactant is \( \ce{H_{2}O(aq)} \) (aqueous water).

Step3: Identify products

For the products, we look at the right side of the arrow. So the products are \( \ce{H_{2}O(l)} \) (liquid water) and \( \ce{O_{2}(g)} \) (gaseous oxygen).

Answer:

• Question 42 (Reactants): \( \ce{H_{2}O(aq)} \)
• Question 43 (Products): \( \ce{H_{2}O(l)} \), \( \ce{O_{2}(g)} \)

For question 39: The \((l)\) next to \( \ce{H_{2}O} \) stands for "liquid". In chemical equations, the state symbol \((l)\) is used to indicate that the substance is in the liquid state.

For question 40: The \((g)\) next to \( \ce{H_{2}O} \) (wait, in the reaction \( \ce{H_{2}O(aq)
ightarrow H_{2}O(l) + O_{2}(g)} \), the \((g)\) is next to \( \ce{O_{2}} \), assuming it's a typo and you mean next to \( \ce{O_{2}} \), the \((g)\) stands for "gas". If it's next to \( \ce{H_{2}O} \) in a different reaction, but in this reaction, for \( \ce{H_{2}O} \) with \((g)\) it would be gaseous water, but in our given reaction, the gaseous substance is \( \ce{O_{2}} \). The \((aq)\) next to \( \ce{H_{2}O} \) stands for "aqueous", meaning the water is dissolved in a solution.

For question 41: The \((s)\) state symbol in chemical equations stands for "solid", indicating that the substance is in the solid state.

For question 31: Let's balance the equation \( \ce{2H_{2} + O_{2}
ightarrow 2H_{2}O} \).

• Step 1: Count atoms on each side
• Left side: \( H\): \( 2\times2 = 4 \), \( O\): \( 2 \)
• Right side: \( H\): \( 2\times2=4 \), \( O\): \( 2\times1 = 2 \)
• Step 2: Check balance

The number of hydrogen and oxygen atoms is the same on both sides. So this equation is balanced. We put a checkmark next to it.

For question 32: Let's balance \( \ce{KCl + Br_{2}
ightarrow KBr + Cl_{2}} \)

• Step 1: Count atoms
• Left side: \( K\): \( 1 \), \( Cl\): \( 1 \), \( Br\): \( 2 \)
• Right side: \( K\): \( 1 \), \( Br\): \( 1 \), \( Cl\): \( 2 \)
• Step 2: Balance the equation

We need to balance \( Br \) and \( Cl \). Multiply \( KCl \) by 2, \( KBr \) by 2. The balanced equation is \( \ce{2KCl + Br_{2}
ightarrow 2KBr + Cl_{2}} \). The original equation is not balanced, so we do not put a checkmark.

For question 33: Let's check \( \ce{Na_{2}CO_{3}+2HCl
ightarrow 2NaCl + H_{2}O + CO_{2}} \)

• Step 1: Count atoms
• Left side: \( Na\): \( 2 \), \( C\): \( 1 \), \( O\): \( 3 + 2\times1=5 \), \( H\): \( 2\times1 = 2 \), \( Cl\): \( 2\times1=2 \)
• Right side: \( Na\): \( 2\times1 = 2 \), \( Cl\): \( 2\times1=2 \), \( H\): \( 2\times1=2 \), \( O\): \( 1+2\times1 = 3 \)? Wait, no. \( H_{2}O \) has 1 O, \( CO_{2} \) has 2 O, total \( 1 + 2=3 \), \( Na_{2}CO_{3} \) has 3 O, \( HCl \) has 0 O (except from \( HCl \) which is \( H^+ \) and \( Cl^- \)). Wait, \( Na_{2}CO_{3}+2HCl

ightarrow 2NaCl + H_{2}O + CO_{2} \)
Left side: \( Na\): 2, \( C\):1, \( O\):3, \( H\):2, \( Cl\):2
Right side: \( Na\):2, \( Cl\):2, \( H\):2, \( O\):1 (from \( H_2O \)) + 2 (from \( CO_2 \))=3, \( C\):1. So it is balanced. Put a checkmark.

For question 34: \( \ce{2Fe_{2}O_{3}+3C
ightarrow 4Fe + 3CO_{2}} \)

• Step 1: Count atoms
• Left side: \( Fe\): \( 2\times2 = 4 \), \( O\): \( 2\times3=6 \), \( C\): \( 3\times1 = 3 \)
• Right side: \( Fe\): \( 4\times1 = 4 \), \( C\): \( 3\times1=3 \), \( O\): \( 3\times2 = 6 \)

It is balanced. Put a checkmark.

For question 35: \( \ce{Na + O_{2}
ightarrow Na_{2}O} \)

• Step 1: Count atoms
• Left side: \( Na\):1, \( O\):2
• Right side: \( Na\):2, \( O\):1

Balance by multiplying \( Na \) by 4, \( O_2 \) by 1, \( Na_2O \) by 2. The balanced equation is \( \ce{4Na + O_{2}
ightarrow 2Na_{2}O} \). Original equation is not balanced, no checkmark.

For question 36: \( \ce{2H_{2}+O_{2}
ightarrow 2H_{2}O} \) (same as question 31). It is balanced, put a checkmark.

For question 37: \( \ce{4H_{2}+2O_{2}
ightarrow 4H_{2}O} \)

• Step 1: Count atoms
• Left side: \( H\): \( 4\times2 = 8 \), \( O\): \( 2\times2 = 4 \)
• Right side: \( H\): \( 4\times2 = 8 \), \( O\): \( 4\times1 = 4 \)

It is balanced (divide all coefficients by 2, we get \( 2H_2+O_2
ightarrow 2H_2O \)). Put a checkmark.

For question 38: \( \ce{2H_{2}+O_{2}
ightarrow H_{2}O} \)

• Step 1: Count atoms
• Left side: \( H\): \( 2\times2 = 4 \), \( O\): \( 2 \)
• Right side: \( H\): \( 2\times1 = 2 \), \( O\): \( 1 \)

Not balanced. Do not put a checkmark.

For question 29: When balancing a chemical reaction, the small numbers (subscripts) in a chemical formula (e.g., the 2 in \( H_2 \)) represent the number of atoms of an element in a molecule. We do not change the subscripts when balancing equations because changing subscripts would change the identity of the substance (e.g., changing \( H_2O \) to \( H_2O_2 \) would make it hydrogen peroxide instead of water). We balance equations by changing coefficients (the big numbers in front of formulas).

For question 30: According to the Law of Conservation of Mass, atoms cannot be destroyed during a reaction. The Law of Conservation of Mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products, which means atoms are neither created nor destroyed, only rearranged.