Question

instructions: balance the chemical equations below.

1. ——n2+——o2→——n2o
2. ——zn+——hcl→——zncl2+——h2
3. ——fe+——o2→——fe2o3
4. ——al+——feo→——al2o3+——fe
5. ——p4+——o2→——p2o5
6. ——c3h8+——o2→——co2+——h2o
7. ——mgf2+——li2co3→——mgco3+——lif
8. ——br2+——h2o→——hbr+——o2
9. ——kbr+——h2so4→——k2so4+——hbr
10. ——c+——so2→——cs2+——co
11. ——al+——h2so4→——al2(so4)3+——h2
12. ——fecl3+——naoh→——fe(oh)3+——nacl

Explanation:

Step1: Balance the first equation

For \(N_2 + O_2 ightarrow N_2O\), balance oxygen atoms. Let the coefficients be \(aN_2 + bO_2 ightarrow cN_2O\).
From nitrogen balance: \(2a = 2c\) (since \(N_2\) has 2 N atoms and \(N_2O\) has 2 N atoms), so \(a = c\).
From oxygen balance: \(2b=c\). Let \(c = 2\), then \(b = 1\) and \(a = 2\). So \(2N_2+O_2 ightarrow 2N_2O\)

Step2: Balance the second equation

For \(Zn + HCl ightarrow ZnCl_2+H_2\), balance Cl atoms. Let the coefficients be \(aZn + bHCl ightarrow cZnCl_2 + dH_2\).
From Zn balance: \(a = c\).
From Cl balance: \(b = 2c\).
From H balance: \(b = 2d\). Let \(c = 1\), then \(a = 1\), \(b = 2\), \(d = 1\). So \(Zn + 2HCl ightarrow ZnCl_2+H_2\)

Step3: Balance the third equation

For \(Fe+O_2 ightarrow Fe_2O_3\), balance O atoms. Let the coefficients be \(aFe + bO_2 ightarrow cFe_2O_3\).
From Fe balance: \(a = 2c\).
From O balance: \(2b = 3c\). Let \(c = 2\), then \(a = 4\), \(b = 3\). So \(4Fe + 3O_2 ightarrow 2Fe_2O_3\)

Step4: Balance the fourth equation

For \(Al+FeO ightarrow Al_2O_3+Fe\), balance O atoms. Let the coefficients be \(aAl + bFeO ightarrow cAl_2O_3 + dFe\).
From O balance: \(b = 3c\).
From Al balance: \(a = 2c\).
From Fe balance: \(b = d\). Let \(c = 1\), then \(a = 2\), \(b = 3\), \(d = 3\). So \(2Al+3FeO ightarrow Al_2O_3 + 3Fe\)

Step5: Balance the fifth equation

For \(P_4+O_2 ightarrow P_2O_5\), balance P atoms. Let the coefficients be \(aP_4 + bO_2 ightarrow cP_2O_5\).
From P balance: \(4a = 2c\).
From O balance: \(2b = 5c\). Let \(a = 1\), then \(c = 2\), \(b = 5\). So \(P_4+5O_2 ightarrow 2P_2O_5\)

Step6: Balance the sixth equation

For \(C_3H_8+O_2 ightarrow CO_2+H_2O\), balance C atoms first. Let the coefficients be \(aC_3H_8 + bO_2 ightarrow cCO_2 + dH_2O\).
From C balance: \(3a = c\).
From H balance: \(8a = 2d\).
From O balance: \(2b = 2c + d\). Let \(a = 1\), then \(c = 3\), \(d = 4\), \(b = 5\). So \(C_3H_8+5O_2 ightarrow 3CO_2 + 4H_2O\)

Step7: Balance the seventh equation

For \(MgF_2+Li_2CO_3 ightarrow MgCO_3+LiF\), balance F atoms. Let the coefficients be \(aMgF_2 + bLi_2CO_3 ightarrow cMgCO_3 + dLiF\).
From Mg balance: \(a = c\).
From F balance: \(2a = d\).
From Li balance: \(2b = d\).
From C and O balance (since \(CO_3\) is a polyatomic ion): \(b = c\). Let \(a = 1\), then \(b = 1\), \(c = 1\), \(d = 2\). So \(MgF_2+Li_2CO_3 ightarrow MgCO_3+2LiF\)

Step8: Balance the eighth equation

For \(Br_2+H_2O ightarrow HBr+O_2\), balance O atoms. Let the coefficients be \(aBr_2 + bH_2O ightarrow cHBr + dO_2\).
From O balance: \(b = 2d\).
From H balance: \(2b = c\).
From Br balance: \(2a = c\). Let \(d = 1\), then \(b = 2\), \(c = 4\), \(a = 2\). So \(2Br_2+2H_2O ightarrow 4HBr+O_2\)

Step9: Balance the ninth equation

For \(KBr+H_2SO_4 ightarrow K_2SO_4+HBr\), balance K atoms. Let the coefficients be \(aKBr + bH_2SO_4 ightarrow cK_2SO_4 + dHBr\).
From K balance: \(a = 2c\).
From Br balance: \(a = d\).
From \(SO_4\) balance (polyatomic ion): \(b = c\). Let \(c = 1\), then \(a = 2\), \(b = 1\), \(d = 2\). So \(2KBr+H_2SO_4 ightarrow K_2SO_4+2HBr\)

Step10: Balance the tenth equation

For \(C+SO_2 ightarrow CS_2+CO\), balance S atoms. Let the coefficients be \(aC + bSO_2 ightarrow cCS_2 + dCO\).
From S balance: \(b = 2c\).
From C balance: \(a = c + d\).
From O balance: \(2b = d\). Let \(c = 1\), then \(b = 2\), \(d = 4\), \(a = 5\). So \(5C+2SO_2 ightarrow CS_2+4CO\)

Step11: Balance the eleventh equation

For \(Al+H_2SO_4 ightarrow Al_2(SO_4)_3+H_2\), balance \(SO_4\) (polyatomic ion). Let the coefficients be \(aAl + bH_2SO_4 ightarrow cAl_2(SO_4)_3 + dH_2\).
From Al balance: \(a = 2c\).
From \(SO_4\) balance: \(b = 3c\).
From H balance: \(2b = 2d\). Let \(c = 1\), then \(a = 2\), \(b = 3\), \(d = 3\). So \(2Al+3H_2SO_4 ightarrow Al_2(SO_4)_3+3H_2\)

Step12: Balance the twelfth equation

For \(FeCl_3+NaOH ightarrow Fe(OH)_3+NaCl\), balance Cl atoms. Let the coefficients be \(aFeCl_3 + bNaOH ightarrow cFe(OH)_3 + dNaCl\).
From Fe balance: \(a = c\).
From Cl balance: \(3a = d\).
From Na balance: \(b = d\).
From O and H balance (since \(OH\) is a polyatomic ion): \(b = 3c\). Let \(a = 1\), then \(b = 3\), \(c = 1\), \(d = 3\). So \(FeCl_3+3NaOH ightarrow Fe(OH)_3+3NaCl\)

Answer:

1. \(2N_2+O_2

ightarrow 2N_2O\)

1. \(Zn + 2HCl

ightarrow ZnCl_2+H_2\)

1. \(4Fe + 3O_2

ightarrow 2Fe_2O_3\)

1. \(2Al+3FeO

ightarrow Al_2O_3 + 3Fe\)

1. \(P_4+5O_2

ightarrow 2P_2O_5\)

1. \(C_3H_8+5O_2

ightarrow 3CO_2 + 4H_2O\)

1. \(MgF_2+Li_2CO_3

ightarrow MgCO_3+2LiF\)

1. \(2Br_2+2H_2O

ightarrow 4HBr+O_2\)

1. \(2KBr+H_2SO_4

ightarrow K_2SO_4+2HBr\)

1. \(5C+2SO_2

ightarrow CS_2+4CO\)

1. \(2Al+3H_2SO_4

ightarrow Al_2(SO_4)_3+3H_2\)

1. \(FeCl_3+3NaOH

ightarrow Fe(OH)_3+3NaCl\)