Question

1. $\ce{\\_\\_\\_c_{8}h_{18} + \\_\\_\\_o_{2}\to\\_\\_\\_co_{2} + \\_\\_\\_h_{2}o}$
2. $\ce{\\_\\_\\_ch_{4} + \\_\\_\\_o_{2}\to\\_\\_\\_co_{2} + \\_\\_\\_h_{2}o}$
3. $\ce{\\_\\_\\_pb(no_{3})_{2} + \\_\\_\\_k_{2}cro_{4}\to\\_\\_\\_pbcro_{4} + \\_\\_\\_kno_{3}}$
4. $\ce{\\_\\_\\_koh + \\_\\_\\_hcl\to\\_\\_\\_kcl + \\_\\_\\_h_{2}o}$
5. $\ce{\\_\\_\\_pb(no_{3})_{2} + \\_\\_\\_ki\to\\_\\_\\_kno_{3} + \\_\\_\\_pbi_{2}}$
6. $\ce{\\_\\_\\_nahco_{3}\to\\_\\_\\_h_{2}o + \\_\\_\\_co_{2} + \\_\\_\\_na_{2}co_{3}}$
7. $\ce{\\_\\_\\_nacl + \\_\\_\\_h_{2}so_{4}\to\\_\\_\\_na_{2}so_{4} + \\_\\_\\_hcl}$
8. $\ce{\\_\\_\\_fes + \\_\\_\\_hcl\to\\_\\_\\_fecl_{2} + \\_\\_\\_h_{2}s}$
9. $\ce{\\_\\_\\_ki + \\_\\_\\_br_{2}\to\\_\\_\\_kbr + \\_\\_\\_i_{2}}$
10. $\ce{\\_\\_\\_nacl + \\_\\_\\_f_{2}\to\\_\\_\\_naf + \\_\\_\\_cl_{2}}$

Explanation:

Step1: Balance the first equation \(C_8H_{18}+O_2

ightarrow CO_2 + H_2O\)
- For carbon: There are 8 carbons in \(C_8H_{18}\). So, if we put 8 in front of \(CO_2\), the carbon atoms are balanced.
- For hydrogen: There are 18 hydrogens in \(C_8H_{18}\). So, if we put 9 in front of \(H_2O\), the hydrogen atoms are balanced.
- For oxygen: On the right - hand side, we have \(8\times2 + 9\times1=25\) oxygen atoms. So, we put \(\frac{25}{2}\) in front of \(O_2\). But to get rid of the fraction, we multiply all coefficients by 2. So the balanced equation is \(2C_8H_{18}+25O_2 = 16CO_2+18H_2O\)

Step2: Balance the second equation \(CH_4+O_2

ightarrow CO_2 + H_2O\)
- For carbon: There is 1 carbon in \(CH_4\), so we put 1 in front of \(CO_2\).
- For hydrogen: There are 4 hydrogens in \(CH_4\), so we put 2 in front of \(H_2O\).
- For oxygen: On the right - hand side, we have \(2 + 2=4\) oxygen atoms. So we put 2 in front of \(O_2\). The balanced equation is \(CH_4 + 2O_2=CO_2+2H_2O\)

Step3: Balance the third equation \(Pb(NO_3)_2+K_2CrO_4

ightarrow PbCrO_4+KNO_3\)
- For lead (\(Pb\)): 1 on each side.
- For chromium (\(Cr\)): 1 on each side.
- For nitrate (\(NO_3\)): There are 2 \(NO_3\) groups in \(Pb(NO_3)_2\). So we put 2 in front of \(KNO_3\).
- For potassium (\(K\)): Since we have 2 \(K\) in \(KNO_3\), we put 1 in front of \(K_2CrO_4\). The balanced equation is \(Pb(NO_3)_2+K_2CrO_4 = PbCrO_4+2KNO_3\)

Step4: Balance the fourth equation \(KOH+HCl

ightarrow KCl + H_2O\)
- For potassium (\(K\)): 1 on each side.
- For chlorine (\(Cl\)): 1 on each side.
- For hydrogen and oxygen: 1 \(H\) and 1 \(O\) on each side. The balanced equation is \(KOH+HCl=KCl + H_2O\)

Step5: Balance the fifth equation \(Pb(NO_3)_2+KI

ightarrow KNO_3+PbI_2\)
- For lead (\(Pb\)): 1 on each side.
- For iodine (\(I\)): There are 2 iodines in \(PbI_2\), so we put 2 in front of \(KI\).
- For nitrate (\(NO_3\)): There are 2 \(NO_3\) groups in \(Pb(NO_3)_2\), so we put 2 in front of \(KNO_3\). The balanced equation is \(Pb(NO_3)_2+2KI = 2KNO_3+PbI_2\)

Step6: Balance the sixth equation \(NaHCO_3

ightarrow H_2O+CO_2+Na_2CO_3\)
- For sodium (\(Na\)): We have 1 on the left and 2 on the right. So we put 2 in front of \(NaHCO_3\).
- For hydrogen: \(2\times1 = 2\) on the left and \(2\times1=2\) on the right (from \(H_2O\)).
- For carbon: \(2\times1 = 2\) on the left (\(2NaHCO_3\)) and \(1 + 1=2\) on the right (\(CO_2+Na_2CO_3\)).
- For oxygen: \(2\times3=6\) on the left and \(1 + 2+3 = 6\) on the right. The balanced equation is \(2NaHCO_3=H_2O+CO_2+Na_2CO_3\)

Step7: Balance the seventh equation \(NaCl+H_2SO_4

ightarrow Na_2SO_4+HCl\)
- For sodium (\(Na\)): We have 1 on the left and 2 on the right. So we put 2 in front of \(NaCl\).
- For chlorine (\(Cl\)): Since we have 2 \(NaCl\), we put 2 in front of \(HCl\). The balanced equation is \(2NaCl+H_2SO_4=Na_2SO_4+2HCl\)

Step8: Balance the eighth equation \(FeS+HCl

ightarrow FeCl_2+H_2S\)
- For iron (\(Fe\)): 1 on each side.
- For sulfur (\(S\)): 1 on each side.
- For chlorine (\(Cl\)): There are 2 chlorines in \(FeCl_2\), so we put 2 in front of \(HCl\). The balanced equation is \(FeS + 2HCl=FeCl_2+H_2S\)

Step9: Balance the ninth equation \(KI+Br_2

ightarrow KBr+I_2\)
- For iodine (\(I\)): There are 2 iodines in \(I_2\), so we put 2 in front of \(KI\).
- For bromine (\(Br\)): There are 2 bromines in \(Br_2\), so we put 2 in front of \(KBr\). The balanced equation is \(2KI+Br_2=2KBr+I_2\)

Step10: Balance the tenth equation \(NaCl+F_2

ightarrow NaF+Cl_2\)
- For chlorine (\(Cl\)): There are 2 chlorines in \(Cl_2\), so we put 2 in front of \(NaCl\).
- For sodium (\(Na\)): Since we have 2 \(NaCl\), we put 2 in front of \(NaF\).
- For fluorine (\(F\)): There are 2 fluorines in \(F_2\) and 2 in \(2NaF\). The balanced equation is \(2NaCl+F_2=2NaF+Cl_2\)

Answer:

1. \(2C_8H_{18}+25O_2 = 16CO_2+18H_2O\)
2. \(CH_4 + 2O_2=CO_2+2H_2O\)
3. \(Pb(NO_3)_2+K_2CrO_4 = PbCrO_4+2KNO_3\)
4. \(KOH+HCl=KCl + H_2O\)
5. \(Pb(NO_3)_2+2KI = 2KNO_3+PbI_2\)
6. \(2NaHCO_3=H_2O+CO_2+Na_2CO_3\)
7. \(2NaCl+H_2SO_4=Na_2SO_4+2HCl\)
8. \(FeS + 2HCl=FeCl_2+H_2S\)
9. \(2KI+Br_2=2KBr+I_2\)
10. \(2NaCl+F_2=2NaF+Cl_2\)