Question

Give the balanced equation for each of the following (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s) If a box is not needed, leave it blank.) a. The combustion of propanol (C,H,OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen b. Aqueous solutions of cadmium(II) nitrate and potassium phosphate are mixed, resulting in the precipitate formation of cadmium(IT) phosphate with aqueous potassium nitrate as the other product c. Solid zinc reacts with aqueous HCl to form aqueous zinc chloride and hydrogen gas d Aqueous barium hydroxide reacts with aqueous hydrobromic acid to produce water and aqueous barium bromide

Answer 1

a)

C3H7OH(l) + O2(g) --> CO2(g) + H2O(g)

balancing equation

consider C

In the reactant side there are 3 C so multiply CO2 by 3 on product side

C3H7OH(l) + O2(g) --> 3CO2(g) + H2O(g)

consider H

In the reactant side there are 8 H so multiply H2O by 4 on product side

C3H7OH(l) + O2(g) --> 3CO2(g) + 4H2O(g)

consider O

In the product side there are 10 O so multiply O2 by 9/2 on reactant side

C3H7OH(l) + 9/2O2(g) --> 3CO2(g) + 4H2O(g)

Multiply by 2 on both sides

2 C3H7OH(l) + 9 O2(g) --> 6 CO2(g) + 8 H2O(g)

b)

Cd(NO3)2(aq) + K3PO4(aq) --> Cd3(PO4)2(s) + KNO3(aq)

balancing equation

consider Cd

In the product side there are 3 Cd so multiply Cd(NO3)2 by 3 on reactant side

3 Cd(NO3)2(aq) + K3PO4(aq) --> Cd3(PO4)2(s) + KNO3(aq)

consider N

In the reactant side there are 6 N so multiply KNO3 by 6 on product side

3 Cd(NO3)2(aq) + K3PO4(aq) --> Cd3(PO4)2(s) + 6 KNO3(aq)

consider K

In the product side there are 6 K so multiply K3PO4 by 2 on reactant side

3 Cd(NO3)2(aq) + 2 K3PO4(aq) --> Cd3(PO4)2(s) + 6 KNO3(aq)

consider O,p

both are balanced in the above equation

so

equation is

3 Cd(NO3)2(aq) + 2 K3PO4(aq) --> Cd3(PO4)2(s) + 6 KNO3(aq)

c)

Zn(s) + HCl(aq) --> ZnCl2(aq) + H2(g)

balancing equation

consider Zn

it is balanced

consider H

In the product side there are 2 H so multiply HCl by 2 on reactant side

Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)

consider Cl

it is balanced in the above equation

so

equation is

Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g

d)

Ba(OH)2 (aq) + HBr(aq) --> H2O(l) + BaBr2(aq)

consider Ba

it is balanced

consider Br

In the product side there are 2 Br so multiply HBr by 2 on reactant side

Ba(OH)2 (aq) + 2 HBr(aq) --> H2O(l) + BaBr2(aq)

consider H

In the reactant side there are 4 H so multiply H2O by 2 on product side

Ba(OH)2 (aq) + 2 HBr(aq) --> 2 H2O(l) + BaBr2(aq)

consider O

it is balanced in the above equation

so

equation is

Ba(OH)2 (aq) + 2 HBr(aq) --> 2 H2O(l) + BaBr2(aq)