Question

balancing equations - #1
a. \\( \frac{4}{1} \mathrm{na}+\frac{1}{2} \mathrm{o}_{2} \
ightarrow \frac{2}{1} \mathrm{na}_{2} \mathrm{o}_{2} \\)
\\( \mathrm{na}=1 \times 4 \\)
\\( \mathrm{o}=2 \times 1 \\)
\\( \mathrm{na}=2 \times 2=4 \\)
\\( \mathrm{o}=1 \times 2=2 \\)
b. \\( \frac{2}{1} \mathrm{h}_{2}+\frac{1}{2} \mathrm{o}_{2} \
ightarrow \frac{2}{1} \mathrm{h}_{2} \mathrm{o} \\)
\\( \mathrm{h}=2 \\)
\\( \mathrm{o}=2 \\)
\\( \mathrm{h}=2 \\)
\\( \mathrm{o}=1 \\)
c. \\( \frac{2}{1} \mathrm{na}_{2} \mathrm{so}_{4}+\frac{1}{2} \mathrm{cacl}_{2} \
ightarrow \frac{1}{1} \mathrm{caso}_{4}+\frac{1}{1} \mathrm{nacl} \\)
\\( \mathrm{na}=2 \\)
\\( \mathrm{ca}=1 \\)
\\( \mathrm{so}=4 \\)
\\( \mathrm{cl}=2 \\)
\\( \mathrm{na}=1 \\)
\\( \mathrm{ca}=1 \\)
\\( \mathrm{so}=4 \\)
\\( \mathrm{cl}=1 \\)
d. \\( \mathrm{al}_{2} \mathrm{o}_{3} \
ightarrow \mathrm{al}+\mathrm{o}_{2} \\)
e. \\( \mathrm{n}_{2}+\mathrm{h}_{2} \
ightarrow \mathrm{nh}_{3} \\)
f. \\( \mathrm{fe}+\mathrm{h}_{2} \mathrm{o} \
ightarrow \mathrm{fe}_{3} \mathrm{o}_{4}+\mathrm{h}_{2} \\)
g. \\( \mathrm{p}_{4}+\mathrm{o}_{2} \
ightarrow \mathrm{p}_{4} \mathrm{o}_{10} \\)
h. \\( \mathrm{c}_{2} \mathrm{h}_{6}+\mathrm{o}_{2} \
ightarrow \mathrm{co}_{2}+\mathrm{h}_{2} \mathrm{o} \\)
i. \\( \mathrm{sicl}_{4} \
ightarrow \mathrm{si}+\mathrm{cl}_{2} \\)

Explanation:

Step1: Balance equation d

For \(Al_2O_3
ightarrow Al + O_2\), balance oxygen first. The least - common multiple of 2 (from \(O_2\)) and 3 (from \(Al_2O_3\)) is 6. So, \(2Al_2O_3
ightarrow Al+3O_2\). Then balance aluminum: \(2Al_2O_3 = 4Al+3O_2\).

Step2: Balance equation e

For \(N_2 + H_2
ightarrow NH_3\), balance nitrogen first. Since there are 2 nitrogen atoms on the left (\(N_2\)), we have \(N_2+H_2
ightarrow 2NH_3\). Then balance hydrogen: \(N_2 + 3H_2=2NH_3\).

Step3: Balance equation f

For \(Fe + H_2O
ightarrow Fe_3O_4+H_2\), balance iron first. Let the coefficient of \(Fe\) be 3. Then for oxygen, since there are 4 oxygen atoms in \(Fe_3O_4\), let the coefficient of \(H_2O\) be 4. For hydrogen, \(4H_2O\) gives 8 hydrogen atoms, so the coefficient of \(H_2\) is 4. The balanced equation is \(3Fe + 4H_2O=Fe_3O_4+4H_2\).

Step4: Balance equation g

For \(P_4+O_2
ightarrow P_4O_{10}\), balance oxygen first. The least - common multiple of 2 (from \(O_2\)) and 10 (from \(P_4O_{10}\)) is 10. So, \(P_4 + 5O_2=P_4O_{10}\).

Step5: Balance equation h

For \(C_2H_6+O_2
ightarrow CO_2+H_2O\), balance carbon first. Let the coefficient of \(CO_2\) be 2 (\(C_2H_6
ightarrow 2CO_2+H_2O\)). Then balance hydrogen. Since there are 6 hydrogen atoms in \(C_2H_6\), let the coefficient of \(H_2O\) be 3 (\(C_2H_6+O_2
ightarrow 2CO_2 + 3H_2O\)). Then balance oxygen: \(C_2H_6+\frac{7}{2}O_2=2CO_2+3H_2O\), multiply through by 2 to get \(2C_2H_6+7O_2 = 4CO_2+6H_2O\).

Step6: Balance equation i

For \(SiCl_4
ightarrow Si+Cl_2\), balance chlorine first. Since there are 4 chlorine atoms in \(SiCl_4\), let the coefficient of \(Cl_2\) be 2. The balanced equation is \(SiCl_4=Si + 2Cl_2\).

Answer:

d. \(2Al_2O_3 = 4Al+3O_2\)
e. \(N_2 + 3H_2=2NH_3\)
f. \(3Fe + 4H_2O=Fe_3O_4+4H_2\)
g. \(P_4 + 5O_2=P_4O_{10}\)
h. \(2C_2H_6+7O_2 = 4CO_2+6H_2O\)
i. \(SiCl_4=Si + 2Cl_2\)