Information on five samples of different compounds are listed below. For each sample, a quantity is

2 answers
Question:

Information on five samples of different compounds are listed below. For each sample, a quantity is given. Complete the table with the quantities of each sample as mass, number of moles, and number of particles for each, and identity for the last sample.
Compound | Mass of sample | # of Moles | # of Particles
BeF2 17.49 g / 0.372 moles / 2.24 x 10^2
1) CH4 / 24.75 g / /
2) MgO / / 0.563 mole /
3) ZnCO3 / / / 1.27 x 10^22
4) PF3 / / 2.25 moles /
5) "Cl2 / 135.92 g / 1.32 moles /

Answers

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.

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.

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