Chemistry 121
Oregon State University
Worksheet 7 Notes
Dr. Barth
1.
In each equation, identify the reactants, the products, and the phase of
each species:
a. S (s) + O2 (g) → SO2 (g)
Reactants:
S (solid)
O2 (gas)
Products:
SO2 (gas)
b. 2 H2O2 (l) → 2 H2O (l) + O2 (g)
Reactants:
H2O2 (liquid)
Products:
H2O (liquid)
O2 (gas)
c. Ca(NO3)2 (s) → Ca2+ (aq) + 2 NO3- (aq)
2.
Reactants:
Ca(NO3)2 (solid)
Products:
Ca2+ (aqueous)
NO3- (aqueous)
Solid sodium metal will react (fairly violently) with water to produce
aqueous sodium hydroxide (NaOH) and hydrogen gas.
a. What are the reactants in this reaction?
Na and H2O
b. What are the products in this reaction?
NaOH and H2
c. What phase is each reactant and product in?
Na (s)
H2O (l)
NaOH (aq)
H2 (g)
d. Write the balanced equation for this reaction
We can balance initially like this:
Na (s) + H2O (l) → NaOH (aq) + ½ H2 (g)
If we want to have whole number coefficients, we multiply all of the coefficients
by the denominator of our fraction (2). That gives:
2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g)
3.
Solid lithium nitrate dissolves in water to produce aqueous lithium and
nitrate ions.
a. What are the reactants in this reaction? Why is water not considered a
reactant?
LiNO3
Water is neither consumed nor produced in this reaction, so it’s not included in
the reaction equation.
b. What are the products in this reaction?
Li+
NO3c. What phase is each reactant and product in?
LiNO3 (s)
Li+ (aq)
NO3- (aq)
d. Write the balanced equation for this reaction
LiNO3 (s) → Li+ (aq) + NO3- (aq)
4.
Balance the following chemical reactions:
a. Cl2O5 + H2O → HClO3
The most complex compound in this reaction is the HClO3 (it doesn’t have the
most atoms, but it does have the most kinds of atoms). We can start by
balancing the atoms in that molecule first.
If we start with H, we can see that there are 2 H on the reactant side, but only
one on the product side. To balance this, we’ll add a stoichiometric coefficient in
front of the HClO3:
Cl2O5 + H2O
→ 2 HClO3
We now have 2 H on each side. This has also served to balance Cl, since we
now have 2 Cl on each side of the reaction arrow. Finally, we can check O. We
have a total of 6 on the reactant side (5 from Cl2O5 and 1 from H2O), and 6 on
the product side. The reaction is now balanced.
b. C2H2 + O2 → CO2 + H2O
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
c. V2O5 + H2
→ V2O3 + H2O
V2O5 + 2 H2
→ V2O3 + 2 H2O
d. Al + ZnO → Al2O3 + Zn
2 Al + 3 ZnO → Al2O3 + 3 Zn