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use with textbook pages 202 - 211. section 4.3 word equations write the…

Question

use with textbook pages 202 - 211. section 4.3
word equations
write the skeleton equation for each of the following reactions. then balance each of the following chemical equations.

  1. hydrogen + oxygen → water
  2. iron(iii) oxide + hydrogen → water + iron
  3. sodium + water → sodium hydroxide + hydrogen
  4. calcium carbide + oxygen → calcium + carbon dioxide
  5. potassium iodide + chlorine → potassium chloride + iodine
  6. chromium + tin(iv) chloride → chromium(iii) chloride + tin
  7. magnesium + copper(ii) sulphate → magnesium sulphate + copper
  8. zinc sulphate + strontium chloride → zinc chloride + strontium sulphate
  9. ammonium chloride + lead(iii) nitrate → ammonium nitrate + lead(iii) chloride
  10. iron(iii) nitrate + magnesium sulphide → iron(iii) sulphide + magnesium nitrate
  11. aluminum chloride + sodium carbonate → aluminum carbonate + sodium chloride
  12. sodium phosphate + calcium hydroxide → sodium hydroxide + calcium phosphate

Explanation:

Step1: Write skeleton equations

For each reaction, write the chemical formulas of reactants and products.

  1. \(H_2+O_2

ightarrow H_2O\)

  1. \(Fe_2O_3 + H_2

ightarrow H_2O+Fe\)

  1. \(Na + H_2O

ightarrow NaOH + H_2\)

  1. \(CaC_2+O_2

ightarrow Ca + CO_2\)

  1. \(KI+Cl_2

ightarrow KCl + I_2\)

  1. \(Cr+SnCl_4

ightarrow CrCl_3+Sn\)

  1. \(Mg+CuSO_4

ightarrow MgSO_4+Cu\)

  1. \(ZnSO_4+SrCl_2

ightarrow ZnCl_2+SrSO_4\)

  1. \(NH_4Cl+Pb(NO_3)_3

ightarrow NH_4NO_3+PbCl_3\)

  1. \(Fe(NO_3)_3+MgS

ightarrow Fe_2S_3+Mg(NO_3)_2\)

  1. \(AlCl_3+Na_2CO_3

ightarrow Al_2(CO_3)_3+NaCl\)

  1. \(Na_3PO_4+Ca(OH)_2

ightarrow NaOH+Ca_3(PO_4)_2\)

Step2: Balance equations

Use the method of balancing atoms one - by - one.

  1. \(2H_2+O_2 = 2H_2O\)
  • Balance \(O\): There are \(2\) \(O\) atoms in \(O_2\) and \(1\) \(O\) atom in \(H_2O\). Multiply \(H_2O\) by \(2\). Then balance \(H\): multiply \(H_2\) by \(2\).
  1. \(Fe_2O_3 + 3H_2=3H_2O + 2Fe\)
  • Balance \(Fe\): \(2\) \(Fe\) atoms in \(Fe_2O_3\), so put \(2\) in front of \(Fe\). Balance \(O\): \(3\) \(O\) atoms in \(Fe_2O_3\), so multiply \(H_2O\) by \(3\). Then balance \(H\): multiply \(H_2\) by \(3\).
  1. \(2Na + 2H_2O=2NaOH + H_2\)
  • Balance \(Na\): multiply \(Na\) and \(NaOH\) by \(2\). Then balance \(H\) and \(O\).
  1. \(CaC_2+2O_2 = Ca + 2CO_2\)
  • Balance \(C\): \(2\) \(C\) atoms in \(CaC_2\), so multiply \(CO_2\) by \(2\). Then balance \(O\): multiply \(O_2\) by \(2\).
  1. \(2KI+Cl_2 = 2KCl+I_2\)
  • Balance \(I\): \(2\) \(I\) atoms in \(KI\) and \(I_2\). Balance \(Cl\): \(2\) \(Cl\) atoms in \(Cl_2\) and \(KCl\).
  1. \(4Cr+3SnCl_4 = 4CrCl_3+3Sn\)
  • Balance \(Cl\): LCM of \(4\) and \(3\) (from \(SnCl_4\) and \(CrCl_3\)) is \(12\). Multiply \(SnCl_4\) by \(3\) and \(CrCl_3\) by \(4\). Then balance \(Cr\) and \(Sn\).
  1. \(Mg+CuSO_4 = MgSO_4+Cu\) (already balanced)
  2. \(ZnSO_4+SrCl_2 = ZnCl_2+SrSO_4\) (already balanced)
  3. \(3NH_4Cl+Pb(NO_3)_3 = 3NH_4NO_3+PbCl_3\)
  • Balance \(Cl\): \(3\) \(Cl\) atoms in \(NH_4Cl\) and \(PbCl_3\). Balance \(NO_3\): \(3\) \(NO_3\) groups.
  1. \(2Fe(NO_3)_3+3MgS = Fe_2S_3+3Mg(NO_3)_2\)
  • Balance \(Fe\): \(2\) \(Fe\) atoms. Balance \(S\): \(3\) \(S\) atoms. Balance \(NO_3\): \(6\) \(NO_3\) groups.
  1. \(2AlCl_3+3Na_2CO_3 = Al_2(CO_3)_3+6NaCl\)
  • Balance \(Al\): \(2\) \(Al\) atoms. Balance \(Na\): \(6\) \(Na\) atoms. Balance \(Cl\) and \(CO_3\).
  1. \(2Na_3PO_4+3Ca(OH)_2 = 6NaOH+Ca_3(PO_4)_2\)
  • Balance \(Na\): \(6\) \(Na\) atoms. Balance \(Ca\): \(3\) \(Ca\) atoms. Balance \(PO_4\) and \(OH\).

Answer:

  1. Skeleton: \(H_2+O_2

ightarrow H_2O\); Balanced: \(2H_2+O_2 = 2H_2O\)

  1. Skeleton: \(Fe_2O_3 + H_2

ightarrow H_2O+Fe\); Balanced: \(Fe_2O_3 + 3H_2=3H_2O + 2Fe\)

  1. Skeleton: \(Na + H_2O

ightarrow NaOH + H_2\); Balanced: \(2Na + 2H_2O=2NaOH + H_2\)

  1. Skeleton: \(CaC_2+O_2

ightarrow Ca + CO_2\); Balanced: \(CaC_2+2O_2 = Ca + 2CO_2\)

  1. Skeleton: \(KI+Cl_2

ightarrow KCl + I_2\); Balanced: \(2KI+Cl_2 = 2KCl+I_2\)

  1. Skeleton: \(Cr+SnCl_4

ightarrow CrCl_3+Sn\); Balanced: \(4Cr+3SnCl_4 = 4CrCl_3+3Sn\)

  1. Skeleton: \(Mg+CuSO_4

ightarrow MgSO_4+Cu\); Balanced: \(Mg+CuSO_4 = MgSO_4+Cu\)

  1. Skeleton: \(ZnSO_4+SrCl_2

ightarrow ZnCl_2+SrSO_4\); Balanced: \(ZnSO_4+SrCl_2 = ZnCl_2+SrSO_4\)

  1. Skeleton: \(NH_4Cl+Pb(NO_3)_3

ightarrow NH_4NO_3+PbCl_3\); Balanced: \(3NH_4Cl+Pb(NO_3)_3 = 3NH_4NO_3+PbCl_3\)

  1. Skeleton: \(Fe(NO_3)_3+MgS

ightarrow Fe_2S_3+Mg(NO_3)_2\); Balanced: \(2Fe(NO_3)_3+3MgS = Fe_2S_3+3Mg(NO_3)_2\)

  1. Skeleton: \(AlCl_3+Na_2CO_3

ightarrow Al_2(CO_3)_3+NaCl\); Balanced: \(2AlCl_3+3Na_2CO_3 = Al_2(CO_3)_3+6NaCl\)

  1. Skeleton: \(Na_3PO_4+Ca(OH)_2

ightarrow NaOH+Ca_3(PO_4)_2\); Balanced: \(2Na_3PO_4+3Ca(OH)_2 = 6NaOH+Ca_3(PO_4)_2\)