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

Step-by-step 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.

Step-by-step 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.