QUESTION IMAGE
Question
balancing chemical equations
suggestions:
- use a pencil, so that you can change coefficients if you make a mistake.
- treat it like a puzzle, not a chore. it can be fun if you look at it as \this is the best mr. glenister can give me. i laugh in the face of easy puzzles like this!\
- write the number of atoms on each side of the equation. if a polyatomic appears on both sides of the equation, treat it as a single atom.
- start with the most complicated compound, as putting a coefficient in front of it will change more atoms; then leave elements until the end, as you can put any coefficient in front of them.
- you can only add coefficients in front of compounds, you cannot change the formula.
- if you need to put a fractional coefficient on something in order to balance it, you can multiply everything by the denominator to clear the fraction.
example:
li + br₂ → libr (br is unbalanced)
1 li 1 li
2 br 1 br
li + br₂ → 2libr (to balance the br, but now li is unbalanced)
1 li 2 li
2 br 2 br
2li + br₂ → 2libr (to balance the li)
2 li 2 li
2 br 2 br
balanced!
ca(no₃)₂ + cu₂so₄ → caso₄ + cuno₃ (cu and no₃ are unbalanced)
1 ca 1 ca
2 no₃ 1 no₃
2 cu 1 cu
1 so₄ 1 so₄
ca(no₃)₂ + cu₂so₄ → caso₄ + 2cuno₃ (to balance cu, which also balanc
1 ca 1 ca
2 no₃ 2 no₃
2 cu 2 cu
1 so₄ 1 so₄
balanced!
Step1: Analyze the unbalanced equation
We are given the un - balanced chemical equation \(Ca(NO_3)_2+Cu_2SO_4
ightarrow CaSO_4 + CuNO_3\). We need to count the number of each type of atom on both sides. On the left - hand side: 1 Ca, 2 \(NO_3\), 2 Cu, 1 \(SO_4\). On the right - hand side: 1 Ca, 1 \(NO_3\), 1 Cu, 1 \(SO_4\). The Cu and \(NO_3\) are unbalanced.
Step2: Balance the Cu atoms
To balance the Cu atoms, we put a coefficient of 2 in front of \(CuNO_3\) on the right - hand side. The equation becomes \(Ca(NO_3)_2+Cu_2SO_4
ightarrow CaSO_4 + 2CuNO_3\).
Step3: Check the balance of other atoms
After adding the coefficient 2 in front of \(CuNO_3\), we check the number of other atoms. For Ca, we have 1 Ca on both sides. For \(NO_3\), we have 2 \(NO_3\) on the left - hand side and \(2\times1 = 2\) \(NO_3\) on the right - hand side. For Cu, we have 2 Cu on both sides. For \(SO_4\), we have 1 \(SO_4\) on both sides.
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\(Ca(NO_3)_2+Cu_2SO_4
ightarrow CaSO_4 + 2CuNO_3\)