Question

ws 6.7 - chemistry problems
#1-5: write the formulas and balance the equation

1. phosphorous + oxygen gas → diphosphorous pentoxide

1. bromine gas + iron → iron (iii) bromide

1. carbon monoxide + iron (iii) oxide → iron +carbon dioxide

1. copper (ii) nitrate + sodium phosphate → copper (ii) phosphate + sodium nitrate

1. bismuth carbonate + lithium acetate → bismuth acetate +lithium carbonate

Explanation:

Problem 1: Phosphorous + Oxygen Gas → Diphosphorous Pentoxide

Step 1: Write the unbalanced formula

Phosphorous is \( P \), Oxygen gas is \( O_2 \), Diphosphorous Pentoxide is \( P_2O_5 \). So the unbalanced equation is \( P + O_2
ightarrow P_2O_5 \).

Step 2: Balance the oxygen atoms

There are 2 O on the left and 5 O on the right. The least common multiple of 2 and 5 is 10. So put a coefficient of 5 in front of \( O_2 \) (to get 10 O) and a coefficient of 2 in front of \( P_2O_5 \) (to get 10 O). Now the equation is \( P + 5O_2
ightarrow 2P_2O_5 \).

Step 3: Balance the phosphorous atoms

Now there are 4 P on the right (from \( 2P_2O_5 \)), so put a coefficient of 4 in front of \( P \) on the left.

Step 1: Write the unbalanced formula

Bromine gas is \( Br_2 \), Iron is \( Fe \), Iron (III) Bromide is \( FeBr_3 \). Unbalanced equation: \( Br_2 + Fe
ightarrow FeBr_3 \).

Step 2: Balance the bromine atoms

There are 2 Br on the left and 3 Br on the right. LCM of 2 and 3 is 6. So put 3 in front of \( Br_2 \) (6 Br) and 2 in front of \( FeBr_3 \) (6 Br). Equation: \( 3Br_2 + Fe
ightarrow 2FeBr_3 \).

Step 3: Balance the iron atoms

Now there are 2 Fe on the right, so put 2 in front of \( Fe \) on the left.

Step 1: Write the unbalanced formula

Carbon Monoxide is \( CO \), Iron (III) Oxide is \( Fe_2O_3 \), Iron is \( Fe \), Carbon Dioxide is \( CO_2 \). Unbalanced equation: \( CO + Fe_2O_3
ightarrow Fe + CO_2 \).

Step 2: Balance the iron atoms

There are 2 Fe in \( Fe_2O_3 \), so put 2 in front of \( Fe \) on the right. Equation: \( CO + Fe_2O_3
ightarrow 2Fe + CO_2 \).

Step 3: Balance the oxygen and carbon atoms

Notice that each \( CO \) becomes \( CO_2 \) by gaining 1 O. \( Fe_2O_3 \) has 3 O to give. So we need 3 \( CO \) to take 3 O (to form 3 \( CO_2 \)). So put 3 in front of \( CO \) and 3 in front of \( CO_2 \).

Answer:

\( 4P + 5O_2
ightarrow 2P_2O_5 \)

Problem 2: Bromine Gas + Iron → Iron (III) Bromide