Question

balancing act
name
atoms are not ____ or ____ during a chemical reaction.
scientists know that there must be the ____ number of atoms on each ____ of
the ____. to balance the chemical equation, you must add ____ in front
of the chemical formulas in the equation. you cannot ____ or ____ subscripts!

1. determine number of atoms for each
2. pick an element that is not equal on

both sides of the equation.

1. add a coefficient in front of the

formula with that element and adjust
your counts.

1. continue adding coefficients to get the

same number of atoms of each element
on each side.
try these:

Explanation:

Step1: Balance the equation \(Ca + O_2

ightarrow CaO\)
Let the coefficients be \(x\), \(y\), \(z\) for \(Ca\), \(O_2\), \(CaO\) respectively. So the equation is \(xCa + yO_2
ightarrow zCaO\).
For \(Ca\): \(x = z\). For \(O\): \(2y=z\). Let \(y = 1\), then \(z = 2\) and \(x=2\). So the balanced equation is \(2Ca+O_2
ightarrow 2CaO\). \(Ca\) (left) \(=2\), \(Ca\) (right) \(=2\), \(O\) (left) \(=2\), \(O\) (right) \(=2\).

Step2: Balance the equation \(N_2 + H_2

ightarrow NH_3\)
Let the coefficients be \(a\), \(b\), \(c\) for \(N_2\), \(H_2\), \(NH_3\) respectively. So the equation is \(aN_2 + bH_2
ightarrow cNH_3\).
For \(N\): \(2a=c\). For \(H\): \(2b = 3c\). Let \(a = 1\), then \(c = 2\). Substitute \(c = 2\) into \(2b=3c\), we get \(2b=6\), \(b = 3\). So the balanced equation is \(N_2+3H_2
ightarrow 2NH_3\). \(N\) (left) \(=2\), \(N\) (right) \(=2\), \(H\) (left) \(=6\), \(H\) (right) \(=6\).

Step3: Balance the equation \(Cu_2O + C

ightarrow Cu+CO_2\)
Let the coefficients be \(m\), \(n\), \(p\), \(q\) for \(Cu_2O\), \(C\), \(Cu\), \(CO_2\) respectively. So the equation is \(mCu_2O + nC
ightarrow pCu+qCO_2\).
For \(Cu\): \(2m=p\). For \(O\): \(m = 2q\). For \(C\): \(n=q\). Let \(q = 1\), then \(m = 2\), \(n = 1\), \(p = 4\). So the balanced equation is \(2Cu_2O + C
ightarrow 4Cu+CO_2\). \(Cu\) (left) \(=4\), \(Cu\) (right) \(=4\), \(O\) (left) \(=2\), \(O\) (right) \(=2\), \(C\) (left) \(=1\), \(C\) (right) \(=1\).

Step4: Balance the equation \(H_2O_2

ightarrow H_2O+O_2\)
Let the coefficient be \(k\) for \(H_2O_2\), \(l\) for \(H_2O\) and \(s\) for \(O_2\). So the equation is \(kH_2O_2
ightarrow lH_2O + sO_2\).
For \(H\): \(2k = 2l\), so \(k=l\). For \(O\): \(2k=l + 2s\). Let \(k = 2\), then \(l = 2\), \(s=1\). So the balanced equation is \(2H_2O_2
ightarrow 2H_2O+O_2\). \(H\) (left) \(=4\), \(H\) (right) \(=4\), \(O\) (left) \(=4\), \(O\) (right) \(=4\).

Answer:

For \(Ca + O_2
ightarrow CaO\): \(Ca\) (left) \(=2\), \(Ca\) (right) \(=2\), \(O\) (left) \(=2\), \(O\) (right) \(=2\).
For \(N_2 + H_2
ightarrow NH_3\): \(N\) (left) \(=2\), \(N\) (right) \(=2\), \(H\) (left) \(=6\), \(H\) (right) \(=6\).
For \(Cu_2O + C
ightarrow Cu+CO_2\): \(Cu\) (left) \(=4\), \(Cu\) (right) \(=4\), \(O\) (left) \(=2\), \(O\) (right) \(=2\), \(C\) (left) \(=1\), \(C\) (right) \(=1\).
For \(H_2O_2
ightarrow H_2O+O_2\): \(H\) (left) \(=4\), \(H\) (right) \(=4\), \(O\) (left) \(=4\), \(O\) (right) \(=4\).