Question

A. Balance the following equations by adding coefficients. Do not leave blank spaces - use a "1" if necessary. Identify the type of reaction in the right column. Balanced Equation Type of Reaction 1. H.As2O7 → _ As2O3 + _H20 2. _N2+_02—_N20 _NaI + _Br2 → _NaBr +_12 PbCrO4 +_HNO3 → __Pb(NO3)2 + __H2CrO4 . _C3H8 +_02 → __CO2 + __H20 TiCl4 + _ Mg → _ MgCl2 + ____Ti CuSO4 +_ KCN → ___Cu(CN)2 +_K2SO4 Ca(ClO3)2 → _ CaCl2 + _ 02 _Na20+ __H_0__ NaOH 10. _CH14 + ___02 → __CO2 + __H20

Answer 1

A. 1.

HYAS O7 As2O5HoO

There are 7 O on left and 6 O on right. So to balance it out write two as coefficient of H₂O.

HNAS2OTA As2 O2H20

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of H₄As₂O₇ and As₂O_5.

_{THAS,OT As2O,+2H»O}

This is the balanced chemical equation. As one reactant (H₄As₂O₇) breaks down to form two products (As₂O₅ and H₂O) so it is a decomposition reaction.

2.

N2O2 N20

There are 2 O on left and 1 O on right. So to balance it out write two as coefficient of N₂O.

O2N2O

There are 2 N on left and 4 N on right, so to balance it out write 2 as coefficient of N₂

$2N_2+O_2 \rightarrow 2N_2O$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of O₂.

$2N_2+1O_2 \rightarrow 2N_2O$

This is the balanced chemical equation. As two reactants (N₂ and O₂) combine to form one product (N₂O) so it is a combination reaction.

3.

NalBr2 NaBr+

There are 2 Br on left and 1 Br on right. Also there are 2 I on right and one on left. So to balance these out write two as coefficient of NaI and NaBr.

$2NaI+Br_2 \rightarrow 2NaBr+I_2$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of I₂ and Br_2.

$2NaI+1Br_2 \rightarrow 2NaBr+1I_2$

This is the balanced chemical equation. As one reactive element (Br₂ in this case) replaces the less reactive element (I₂ in this case) from its compound (NaI in this case), so this is a single replacement reaction.

4. $PbCrO_4+HNO_3 \rightarrow Pb(NO_3)_2+H_2CrO_4$

There are 2 H on right and 1 H on left. There are 2 NO₃ groups on right and one on left. So to balance these out write two as coefficient of HNO₃.

$PbCrO_4+2HNO_3 \rightarrow Pb(NO_3)_2+H_2CrO_4$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of PbCrO₄ and H₂CrO_4.

$1PbCrO_4+2HNO_3 \rightarrow Pb(NO_3)_2+1H_2CrO_4$

This is the balanced chemical equation. As there is a mutual exchange of ions between the two reacting compounds to from two new compounds this is a double replacement reaction.

5. $C_3H_8+O_2 \rightarrow CO_2+H_2O$

There are 3 C on left and one on right. So to balance it out write three as coefficient of CO₂.

$C_3H_8+O_2 \rightarrow 3CO_2+H_2O$

There are 8 H on left and 2 on right so to balance it out write 4 as coefficient of H₂O.

$C_3H_8+O_2 \rightarrow 3CO_2+4H_2O$

Now there are 3x2+4=6+4=10 O on right and 2 O on left. So to balance it out write 5 as coefficient of O₂

$C_3H_8+5O_2 \rightarrow 3CO_2+4H_2O$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of C₃H₈

$1C_3H_8+5O_2 \rightarrow 3CO_2+4H_2O$

This is the balanced chemical equation. As in this reaction a hydrocarbon (C₃H₈) reacts with oxygen (O₂) to form CO₂ and H₂O, this is a combustion reaction.

6. $TiCl_4+Mg \rightarrow MgCl_2+Ti$

There are 4 Cl on left and 2 Cl on right. So to balance it out write two as coefficient of MgCl₂.

$TiCl_4+Mg \rightarrow 2MgCl_2+Ti$

There are 2 Mg on right and one on left, so to balance it out write 2 as coefficient of Mg.

$TiCl_4+2Mg \rightarrow 2MgCl_2+Ti$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of TiCl₄ and Ti_.

$1TiCl_4+2Mg \rightarrow 2MgCl_2+1Ti$

This is the balanced chemical equation. As one reactive element (Mg in this case) replaces the less reactive element (Ti in this case) from its compound (TiCl₄ in this case), so this is a single replacement reaction.

7. $CuSO_4+KCN \rightarrow Cu(CN)_2+K_2SO_4$

There are 2 CN on right and 1 on left. So to balance it out write two as coefficient of KCN.

$CuSO_4+2KCN \rightarrow Cu(CN)_2+K_2SO_4$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of CuSO₄ and K₂SO_4.

$1CuSO_4+2KCN \rightarrow Cu(CN)_2+1K_2SO_4$

This is the balanced chemical equation. As there is a mutual exchange of ions between the two reacting compounds to from two new compounds this is a double replacement reaction.

8. $Ca(ClO_3)_2 \rightarrow CaCl_2+O_2$

There are 3x2=6 O on left and 2 O on right. So to balance it out write three as coefficient of O₂.

$Ca(ClO_3)_2 \rightarrow CaCl_2+3O_2$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of Ca(ClO₃)₂ and CaCl_2.

$1Ca(ClO_3)_2 \rightarrow 1CaCl_2+3O_2$

This is the balanced chemical equation. As one reactant (Ca(ClO₃)₂) breaks down to form two products (CaCl₂ and O₂) so it is a decomposition reaction.

9. $Na_2O +H_2O \rightarrow NaOH$

There are 2 Na on left and 1 O on right. So to balance it out write two as coefficient of NaOH.

$Na_2O +H_2O \rightarrow 2NaOH$

Now we can see that all the atoms are balanced in this reaction. Write 1 as coefficient of Na₂O and H₂O_.

$1Na_2O +1H_2O \rightarrow 2NaOH$

This is the balanced chemical equation. As two reactants (Na₂O and H₂O) combine to form one product (NaOH) so it is a combination reaction.

10. $C_6H_{14}+O_2 \rightarrow CO_2+H_2O$

There are 6 C on left and one on right. So to balance it out write six as coefficient of CO₂.

$C_6H_{14}+O_2 \rightarrow 6CO_2+H_2O$

There are 14 H on left and 2 on right so to balance it out write 7 as coefficient of H₂O.

$C_6H_{14}+O_2 \rightarrow 6CO_2+7H_2O$

Now there are 6x2+7=12+7=19 O on right and 2 O on left. So to balance it out write 19/2 as coefficient of O₂

$C_6H_{14}+\frac{19}{2}O_2 \rightarrow 6CO_2+7H_2O$

Multiply the whole equation by two

$2C_6H_{14}+19O_2 \rightarrow 12CO_2+14H_2O$

Now we can see that all the atoms are balanced in this reaction.

This is the balanced chemical equation. As in this reaction a hydrocarbon (C₆H₁₄) reacts with oxygen (O₂) to form CO₂ and H₂O, this is a combustion reaction.