Question

1. ____ no_{2} + ____ h_{2}o \to ____ nh_{3} + ____ o_{2}

Explanation:

Step1: Assign variables to coefficients

Let the coefficients be \(a\), \(b\), \(c\), \(d\) for \(NO_2\), \(H_2O\), \(NH_3\), \(O_2\) respectively. So the equation is \(aNO_2 + bH_2O
ightarrow cNH_3 + dO_2\).

Step2: Balance nitrogen atoms

The number of nitrogen atoms on the left - hand side is \(a\), and on the right - hand side is \(c\). So \(a = c\).

Step3: Balance hydrogen atoms

The number of hydrogen atoms on the left - hand side is \(2b\), and on the right - hand side is \(3c\). Since \(a = c\), we have \(2b=3a\).

Step4: Balance oxygen atoms

The number of oxygen atoms on the left - hand side is \(2a + b\), and on the right - hand side is \(2d\). Substitute \(b=\frac{3}{2}a\) into \(2a + b=2d\), we get \(2a+\frac{3}{2}a = 2d\), which simplifies to \(\frac{4a + 3a}{2}=2d\), or \(\frac{7a}{2}=2d\), or \(d=\frac{7}{4}a\).

Step5: Choose \(a = 4\) (to get rid of fractions)

If \(a = 4\), then \(c=a = 4\), \(b=\frac{3}{2}a=6\), \(d=\frac{7}{4}a = 7\).

Answer:

\(4NO_2+6H_2O
ightarrow4NH_3 + 7O_2\)