Question

art d: practice problems - balance each equation using the process from part c.
cl₂ + nabr → nacl + br₂
na + br₂ → nabr
hgo + cl₂ → hgcl + o₂
h₂ + n₂ → nh₃
cucl₂ + h₂s → cus + hcl
c + h₂ → ch₄
challenge problem: give it your best shot!
c₂h₆ + o₂ → co₂ + h₂o

Explanation:

Step1: Balance the carbon atoms

For the equation \(C_{2}H_{6}+O_{2}\to CO_{2}+H_{2}O\), there are 2 carbon atoms on the left - hand side (\(C_{2}H_{6}\)). So, we put a coefficient of 2 in front of \(CO_{2}\) on the right - hand side. The equation becomes \(C_{2}H_{6}+O_{2}\to 2CO_{2}+H_{2}O\).

Step2: Balance the hydrogen atoms

There are 6 hydrogen atoms in \(C_{2}H_{6}\). So, we put a coefficient of 3 in front of \(H_{2}O\) (since \(3\times2 = 6\) hydrogen atoms). The equation is now \(C_{2}H_{6}+O_{2}\to 2CO_{2}+3H_{2}O\).

Step3: Balance the oxygen atoms

On the right - hand side, the number of oxygen atoms is \(2\times2+3\times1=4 + 3=7\) (from \(2CO_{2}\) and \(3H_{2}O\)). On the left - hand side, we have \(O_{2}\). We can write the coefficient of \(O_{2}\) as \(\frac{7}{2}\), but we usually want whole - number coefficients. So, we multiply the entire equation by 2 to get rid of the fraction.

Multiplying each term by 2: \(2C_{2}H_{6}+7O_{2}\to 4CO_{2}+6H_{2}O\).

Answer:

\(2C_{2}H_{6}+7O_{2}\to 4CO_{2}+6H_{2}O\)