Chapter 12
Study Guide - Answers
1.
What is the Law of Conservation of Matter?
Matter can be neither created nor destroyed in a chemical reaction.
2.
Convert 56L of sulfur dioxide gas into grams of sulfur dioxide gas.
56 L SO2
3.
1 mol SO2
x
22.4 L SO2
x
64.064 g SO2
1 mol SO2
=
160.16 g SO2
What would the volume of 1 mole of carbon dioxide be at STP?
22.4 L
4.
What are the values of standard temperature and pressure?
0oC and 1 atm
5.
Balance the following equations:
C6H12(l) + 9O2(g) 6CO2(g) + 6H2O
Ba(NO3)2(aq) + Na2CrO4(aq) BaCrO4(s) + 2NaNO3(aq)
6.
According to the following unbalanced equation, how many moles of H3PO4 would be required to react with
3 moles of BaCl2? (2 moles)
3BaCl2(aq) + 2H3PO4(aq)  Ba3(PO4)2(s) + 6HCl(aq)
a. How many moles of H3PO4 would be required to react with 4 moles of BaCl2? (2.666 moles)
b. How many moles of H3PO4 would be required to react with 5 moles of BaCl2? (3.33 moles) If 5 moles of
BaCl2 did react with this amount of H3PO4, how many moles of Ba3(PO4)2 would be produced? (1.667
moles)
7.
Given the equation K2PtCl4 + NH3  KCl + PtCl2(NH3)2, find the mass of PtCl2(NH3)2 formed if 300g of
NH3 reacts.
K2PtCl4 + 2NH3  2KCl + PtCl2(NH3)2
300 g NH3
x
1 mol NH3
17.031 g NH3
x
1 mol PtCl2(NH3)2
2 mol NH3
x
300.048 g PtCl2(NH3)2
1 mol PtCl2(NH3)2
=
2643 g
PtCl2(NH3)2
Chapter 12
Study Guide - Answers
8.
Given the equation RbCl + O2  RbClO4, find the mass of RbClO4 formed if 16L of O2 reacts with 155g
RbCl.
RbCl + 2O2  RbClO4
16 L O2
155 g RbCl
1 mol O2
x
22.4 L O2
x
1 mol RbClO4
x
1 mol RbCl
2 mol O2
x
120.921 g RbCl
1 mol RbClO4
1 mol RbCl
184.917 g RbClO4
x
1 mol RbClO4
184.917 g RbClO4
x
1 mol RbClO4
66 g RbClO4
=
=
237 g RbClO4
Find the amount of non-limiting reactant left over after the reaction.
(Find the amount of RbCl that would react with 16L of O2.)
16 L O2
1 mol O2
x
1 mol RbCl
x
22.4 L O2
2 mol O2
120.921 g RbCl
x
1 mol RbCl
43.19 g RbCl
=
(Subtract the amount of RbCl reacted from that supplied.)
155 g RbCl – 43.19 g RbCl = 111.81 g RbCl remaining.
9.
Given the unbalanced equation Ba(C2H3O2)2 + Na3PO4  Ba3(PO4)2 + NaC2H3O2, find the mass of
Ba3(PO4)2 formed if 3.4 x 1025 formula units of Ba(C2H3O2)2 reacts.
3Ba(C2H3O2)2 + 2Na3PO4  Ba3(PO4)2 + 6NaC2H3O2,
3.4 x 1025 f.u.
Ba(C2H3O2)2
1 mol
x
Ba(C2H3O2)2
23
6.02 x 10 f.u.
Ba(C2H3O2)2
x
1 mol Ba3(PO4)2
3 mol
Ba(C2H3O2)2
x
601.93 g Ba3(PO4)2
1 mol Ba3(PO4)2
=
11332 g
Ba3(PO4)2