Balance each equation by inserting coefficients as needed. equation 1: C_{3}H_{8} + O_{2} -> CO_{2} + H_{2}O C3H8+O2⟶CO2+H2O equation 2: P_{4}O_{10} + H_{2}O -> H_{3}PO_{4} P4O10+H2O⟶H3PO4

Answer 1

Hello, Jaylin,

The question is difficult to read. I'll assume you want to balance two equations:

1. C3H8+O2⟶CO2+H2O

Start with the most difficult molecule. I'll choose C₃H₈. It contributes 3 carbons to the product side, so let's add a 3 in front of the CO₂.

C3H8+O2⟶3CO2+H2O

It also contributes 8 hydrogens, so let's add a coefficient of 4 to the H₂O.

C3H8+O2⟶3CO2+4H2O

The carbons and hydrogens now balance (same number on both sides). Now let's address the oxygens.

There are a total of 10 oxygens on the product side (6 from 2CO₂, and 4 from 4H2O). Let's make the coefficient for O₂ 5, since that will provide all the oxygens we need.

Thus: C3H8+5O2⟶3CO2+4H2O is now a balanced equation.

1. Use the same procedure for the second problem:

P₄O₁₀+H₂O⟶H₃PO₄

1. P: P₄O₁₀+H₂O⟶4H₃PO₄ ---- 4 P on each side
2. O: P₄O₁₀+6H₂O⟶4H₃PO₄ ---- 16 O on each side
3. H: P₄O₁₀+6H₂O⟶4H₃PO₄ ---- 12 H on each side

P₄O₁₀+6H₂O⟶4H₃PO₄ is balanced



Fun,

Bob