Balance the following equations using the method outlined in Section 3.7 : (a) \(\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}\) (b) \(\mathrm{CO}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) (c) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \longrightarrow \mathrm{HBr}\) (d) \(\mathrm{K}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{KOH}+\mathrm{H}_{2}\) (e) \(\mathrm{Mg}+\mathrm{O}_{2} \longrightarrow \mathrm{MgO}\) (f) \(\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}\) (g) \(\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (h) \(\mathrm{N}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{NH}_{3}\) (i) \(\mathrm{Zn}+\mathrm{AgCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{Ag}\) (j) \(\mathrm{S}_{8}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2}\) (k) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (l) \(\mathrm{Cl}_{2}+\mathrm{NaI} \longrightarrow \mathrm{NaCl}+\mathrm{I}_{2}\) \((\mathrm{m}) \mathrm{KOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{CH}_{4}+\mathrm{Br}_{2} \longrightarrow \mathrm{CBr}_{4}+\mathrm{HBr}\)

Short Answer

Expert verified
The balanced equations are: (a) \(2\mathrm{C}+ \mathrm{O}_{2} \longrightarrow 2\mathrm{CO}\) (b) \(\mathrm{CO}+ \frac{1}{2}\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) (c) \(\mathrm{H}_{2}+ \mathrm{Br}_{2} \longrightarrow 2\mathrm{HBr}\) (d) \(2\mathrm{K}+2\mathrm{H}_{2} \mathrm{O} \longrightarrow 2\mathrm{KOH}+ \mathrm{H}_{2}\) (e) \(2\mathrm{Mg}+ \mathrm{O}_{2} \longrightarrow 2\mathrm{MgO}\) (f) \(2\mathrm{O}_{3} \longrightarrow 3\mathrm{O}_{2}\) (g) \(2\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (h) \(\mathrm{N}_{2}+3\mathrm{H}_{2} \longrightarrow 2\mathrm{NH}_{3}\) (i) \(\mathrm{Zn}+2\mathrm{AgCl} \longrightarrow \mathrm{ZnCl}_{2}+2\mathrm{Ag}\) (j) \(S_{8}+8\mathrm{O}_{2} \longrightarrow 8\mathrm{SO}_{2}\) (k) \(2\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2\mathrm{H}_{2} \mathrm{O}\) (l) \(\mathrm{Cl}_{2}+2\mathrm{NaI} \longrightarrow 2\mathrm{NaCl}+\mathrm{I}_{2}\) (m) \(\mathrm{KOH}+ \frac{1}{3}\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \frac{1}{3}\mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{CH}_{4}+\mathrm{Br}_{2} \longrightarrow \mathrm{CBr}_{4}+4\mathrm{HBr}\)

Step by step solution

01

Analyzing the individual equations

It's important to notice that each equation has different elements in the reactants and products, however, the basic principle of balancing remains the same - ensuring that the count of each element on both sides of the equation is equal.
02

Balancing the equations

Let's balance each equation individually, keeping in mind that we can only change the coefficients in front of the chemical compounds, not the subscripts within them. (a) \(2\mathrm{C}+ \mathrm{O}_{2} \longrightarrow 2\mathrm{CO}\) (b) \(\mathrm{CO}+ \frac{1}{2}\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}\) (c) \(\mathrm{H}_{2}+ \mathrm{Br}_{2} \longrightarrow 2\mathrm{HBr}\) (d) \(2\mathrm{K}+2\mathrm{H}_{2} \mathrm{O} \longrightarrow 2\mathrm{KOH}+ \mathrm{H}_{2}\) (e) \(2\mathrm{Mg}+ \mathrm{O}_{2} \longrightarrow 2\mathrm{MgO}\) (f) \(2\mathrm{O}_{3} \longrightarrow 3\mathrm{O}_{2}\) (g) \(2\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (h) \(\mathrm{N}_{2}+3\mathrm{H}_{2} \longrightarrow 2\mathrm{NH}_{3}\) (i) \(\mathrm{Zn}+2\mathrm{AgCl} \longrightarrow \mathrm{ZnCl}_{2}+2\mathrm{Ag}\) (j) \(S_{8}+8\mathrm{O}_{2} \longrightarrow 8\mathrm{SO}_{2}\) (k) \(2\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2\mathrm{H}_{2} \mathrm{O}\) (l) \(\mathrm{Cl}_{2}+2\mathrm{NaI} \longrightarrow 2\mathrm{NaCl}+\mathrm{I}_{2}\) (m) \(\mathrm{KOH}+ \frac{1}{3}\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \frac{1}{3}\mathrm{K}_{3} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (n) \(\mathrm{CH}_{4}+\mathrm{Br}_{2} \longrightarrow \mathrm{CBr}_{4}+4\mathrm{HBr}\)

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