Question

PLEASE HELP ME PLEASE PLEASE PLEASE
Given the reaction, PCl3 + Cl2 Imported Asset PCl5, if 3.00 moles of Cl2 are used, then how many moles of PCl5 are made?
6.00 moles
5.00 moles
3.00 moles
1.00 mole

Given the reaction, Fe2O3 + 3 CO Imported Asset 2 Fe + 3 CO2, if 12.0 moles of CO2 are made, then how many moles of CO were used?
12.0 moles
6.0 moles
4.0 moles
3.0 moles

Given the reaction, 2 SO2 + O2 Imported Asset 2 SO3, if 4 moles of SO2 are used, then how many moles of SO3 are made?
4 moles
2 moles
1 mole
1/2 mole

Using the equation, 4Fe + 3O2 Imported Asset 2Fe2O3, if 6 moles of oxygen and an excess of iron were available, how many moles of iron (III) oxide would be produced?
3 moles
4 moles
5 moles
6 moles

How much volume would 2.00 moles of gas take up, under standard temperature and pressure conditions?
11.2 L
22.4 L
33.6 L
44.8 L

If you have 3.0 liters of an unknown gas, at standard temperature and pressure conditions, then you can calculate what about that gas?
mass
density
moles of representative particles
mass and moles of representative particles

When magnesium metal and an aqueous solution of hydrochloric acid combine, they produce an aqueous solution of magnesium chloride and hydrogen gas. Using the equation, Mg (s) + 2HCl (aq) Imported Asset MgCl2 (aq) + H2 (g), if 48.6 g of Mg and 150.0 g of HCl are allowed to react, identify the limiting reagent.
Mg
HCl
MgCl2
H2

When magnesium metal and an aqueous solution of hydrochloric acid combine, they produce an aqueous solution of magnesium chloride and hydrogen gas. Using the equation, Mg (s) + 2HCl (aq) Imported Asset MgCl2 (aq) + H2 (g), if 24.3 g of Mg and 75.0 g of HCl are allowed to react, calculate the mass of H2 that is produced.
1.00 g
2.00 g
4.00 g
5.00 g

The limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product.
TRUE
FALSE

Ammonia gas is formed from nitrogen gas and hydrogen gas, according to the following equation: N2 (g) + 3H2 (g) Imported Asset 2NH3 (g). If 84.0 g of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 85.0 g of ammonia, what is the percent yield of this reaction?
42.2%
65.0%
70.3%
83.3%

Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) Imported Asset Hg (l) + O2 (g). If 3.55 moles of HgO decompose to form 1.54 moles of O2 and 618 g of Hg, what is the percent yield of this reaction?
13.2%
42.5%
56.6%
86.5%

Ammonia gas is formed from nitrogen gas and hydrogen gas according to the following equation: N2 (g) + 3H2 (g) Imported Asset 2NH3 (g). If 112 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 120 grams of ammonia, what is the percent yield of this reaction?
44.1%
66.2%
88.3%
96.4%

Once the percent yield has been determined for a reaction, that percent yield will never vary.
TRUE
FALSE

The theoretical yield for a chemical reaction can not be calculated until the reaction is completed.

TRUE
FALSE

Answer 1

1) A 
"arrow" should be "yields." 
The coefficients are mole ratios. So every 4 moles of NH3, or ammonia, produce 6 moles of H2O, water. 

(12 mol NH3)(6 mol H2O/4 mol NH3) = 18 mol H2O produced 

2) C 
Again, use the coefficients to form mole ratios and solve. See #1. 

3) C 
4)C 
5) C 
Moles, clearly. It's another constant you MUST MEMORIZE, like Avogadro's number. (Shame on you if you don't know what Avogadro's number means!) At STP (273.15 K and 1 atm), one mole of any ideal gas is 22.4 L. 

6) C 
Same thing as #1-4, except with a twist: molar masses. You calculate the molar masses using the periodic table. 

Molar mass of H = 1.008 g/mol 
Molar mass of O = 16.00 g/mol 
Molar mass of H2O = 2(1.008 g/mol) + 16.00 g/mol) = 18.016 g/mol 
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol 

Divide the mass of water by the molar mass to find the number of moles of water. Use mole ratios to find the number of moles of O2 consumed. Then, multiply by the molar mass to find mass of O2. Tada! 

7) 
8) 
9) 
10) 
11) 
12) See number 13 
13) True 
14) False; doesn't tell a thing about the speed of the reaction, how spontaneous it is, or at what temperature or pressure it must take place.

Answer 2

Question 1

The  moles of  PCl5  is    3  moles

calculation

PCl3  +Cl2 →PCl5

from equation above the  mole  ratio  of  cl2 :PCl5 is  1:1  therefore the  moles  of PCl5 is also= 3.00 moles ( answer C)

question 2

The   moles of Co that are made is   12  moles

calculation

Fe2O3  +3Co →2Fe  + 3CO2

from equation above  the  mole ratio  of CO:CO2  is 3:3 therefore the   moles of CO  =  12.0  moles  x 3/3  = 12  moles

question 3

 The  moles of SO3  is 4 moles

calculation

2SO2  +O2→ 2SO3

from  equation above  the  mole ratio of SO2:SO3   is  2:2 therefore the  moles  of SO3

 = 4  x2/2  = 4 moles

question 4

The   moles of  iron (iii) oxide  is  4 moles

calculation

4Fe +3O2 → 2Fe2O3

from equation above  the moles of O2 :Fe2O3 is 3:2  therefore  the moles of Fe2O3  is  

    =  6  moles x 2/3 =  4  moles

         

Question  5

The  volume  2.00  moles  of gas take up  is  44.8 L

calculation

At STP   1   mole= 22.4 L

              2  moles =?

by  cross multiplication  =( 2  moles x 22.4 L) /1 mole =44.8 L

Question  6

3.0  liters of unknown  gas at STP can calculate moles  of   representative  particles

  Explanation

At STP   1 mole =  22.4  l

                  ? moles   =   3 L

by cross  multiplication  = ( 3L x 1 mole) / 22.4 L=0.134  moles

Therefore  3.0 L  of unknown gas  can be used to calculate  the moles of  representative  particles

Question 7

The limiting reagent  is  Mg

Explanation

Mg +2HCl → MgCl2 +H2

calculate  the  moles  of each reactant

moles=mass/molar  mass

moles   of  Mg= 48.6 g/24.3 =2  moles

moles  of HCl = 150.0/36.5 = 4.11 moles

The  moles ratio of Mg:MgCl2  is 1:1 therefore Mg reacted to produce 2  moles  of MgCl2

The  mole ratio  of HCl :MgCl2  is 2:1 therefore the HCl reacted to produce 4.11 x1/2 =2.055 moles  of MgCl2

since  Mg  is  total consumed Mg  is the  limiting reagent

Question 8

The mass  of H2 produced  is  2.00 g

 calculation

Mg  +2 HCl → MgCl2 +H2

calculate  the moles  of  each  reactant

moles  of  mg = 24.3 g/24.3 g/mol=  1  mole

              for H2 = 75.0 g/ 2  g/mol =37.5  moles

Mg is  the limiting  reagent

The mole ratio  of Mg:H2 is 1:1 therefore the  moles of H2 = 1 mole

mass= moles x molar mass

= 1 mole x 2 g/mol   = 2.00 g

Question  9

it is true   that   the limiting reagent  is the reactant  that  is completely  consumed  in a  chemical  reaction and  limits  the  amount  of product.

question 10

The %  yield  is  83.3%

calculation

% yield  =actual yield  /theoretical yield  x 100

actual yield =  85.0 g

calculate the  theoretical  yield  as below

N2(g)  +3 H2 → 2NH3

find  the  moles  of N2  = mass/molar mass

moles of  N2  =  84.0g /28g/mol=  3  moles

from equation above the  mole ratio  of N2: NH3  is 1:2  therefore the moles of NH3  

=3 mole x2  =  6 moles

Theoretical  mass = moles x molar mass

= 6 moles  x 17 g/mol = 102 g

% yield  is therefore  = 85 g 102 g  x 100 =83.3%

Question 11

The  %  yield is  86.6%

calculation

% yield  =  actual  moles/theoretical moles x 100

theoretical moles =   3.55  moles

calculate the  actual  moles  as below

moles= mass/molar mass

=  618  g /201 g/mol  =3.075  moles

% yield  = 3.075 /3.55  x 100  =  86.6%

Question  12

The % yield  of this   reaction is  88.2%

calculation

% yield =  actual  mass/theoretical  mass x 100

actual mass = 120  g

 calculate the  theoretical  mass  as below

N2  +3 H2 → 2NH3

find the  moles of N2 =  mass/molar mass

=112 g  /28  = 4 moles

from the equation above the  mole  ratio  of N2:NH3 is 1:2  rherefore  the  moles of NH3  

 =  4 moles x2  =  8 moles

the theoretical mass of NH3  is =  moles  x molar  mass

=  8  moles x 1 7 g /mol = 136  g

%  yield =  120 g/136 g  x100 =88.2%

question  13

False that  once the percent  yield has  been   determined for  reaction,  that  percent yield  will never  vary.

Explanation

The percent  yield  may  vary  due  to  other external factors  such  as temperature  which  affect chemical equation

question 14

false  The   theoretical  yield  for  a  chemical  reaction  can be calculated  before  the  reaction  is  complete .

 The theoretical yield can be calculated by  determining  the  expected  ratio of number of moles of  limiting  reactant.