6 10. Complete the table below: WATCH FOR FRACTIONS mol H:OExcess molCH Excess m

Question

6 10. Complete the table below: WATCH FOR FRACTIONS mol H:OExcess molCH Excess mol O2 4 molClick here Click here to Click here to enter Click here to enter to enter enter text tect text 3 mol enter text text text to enter text 2 mol Click here Click here Cick here click here to enter to enter to enter text to enter text text Click here Click here 3 mol Click here to Click here to enterClick here to enter to enterto enter enter text text text Basic Stoichiometry Post-Lab Exereises Complete each exercise. Remember to use proper units and labels. 1. For the reaction-N. +-0, _NO·determine the correct lowest mole ratio. Click here to enter text 2. For the reaction-so, +-o, -so, determine the correct lowest mole ratio. Click here to enter text 3 For the reaction P.+6C, 4PCI,, determine how many moles of chlorine Ch would be needed to react with 3 moles of phosphorus P4 to entirely use up all the phosphorus 4) Click here to enter text 4. If 5 moles of P. reacted with 22 moles Clz according to the above Auction, determine: a) How many moles PCh are produced b) How many moles of Pi are left in excess after the reaction (if any) b) Click here to enter text c) How many moles of Cla are left in excess after the reaction (if any)c) Click here to enter text a) Click here to enter text Print Layout View Sec1 Pages: 4 of 5 Words: 545 of 1057 100%

Explanation / Answer

Answer:

10. The reaction is CH4 + 2O2 >> CO2 + 2H2O

a) Moles of CH4 = 4 mol

Moles of O2 = 4 mol

Divide by the co-efficients of balanced equation

CH4 >> 4/1 = 4 mol

O2 >> 4/2 = 2 mol

Oxygen is the lower value, So Oxygen is the limiting reagent.

i) O2 : CO2 molar ratio from balanced equation

2 : 1

4 : x

2x = 4 >> x = 2 mol of CO2

Moles of CO2 = 2 mol

ii) O2 : H2O molar ratio from balanced equation

2 : 2

4 : x

2x = 8 >> x = 4 mol

Therefore, moles of H2O = 4 mol

iii) Excess mol of CH4:

CH4 : O2 molar ratio from balanced equation

1 : 2

x : 4

2x = 4 >> x = 2

Moles of CH4 = 2 mol

Excess mol of CH4 = 4 mol - 2 mol = 2 mol

iv) Excess mol of O2 :

It is a limiting reagent. So, there is no oxygen left behind.

b) Moles of CH4 = 3 mol

Moles of O2 = 6 mol

To find the limiting reagent, divide by co-efficients of balanced equation

CH4 >> 3/1 > 3 mol

O2 >> 6/2 > 3 mol

So, both CH4 and O2 are having same number of moles. There is no excess reagent.Both are limiting reagent.

i) Find the moles of CO2

CH4 : CO2 molar ratio

1 : 1

3 : x

x = 3 mol

Moles of CO2 = 3 mol

ii) Find the moles of H2O

CH4 : H2O molar ratio

1 : 2

3 : x

x = 6 mol

Moles of H2O = 6 mol

iii) Excess mol of CH4:

Its zero. Because it is a limiting reagent. All the moles of CH4 consumed.

iv) Excess mol of O2:

Its zero. Because it is a limiting reagent. All the moles of O2 consumed in the reaction.

c) Given,

Moles of CO2 = 2 mol

Moles of H2O = 4 mol

To find the limiting reagent, Divide by the co-efficients of the balanced equation

CO2 >> 2/1 = 2 mol

H2O >> 4/2 = 2 mol   

So, both CO2 and H2O are having same number of moles. There is no excess reagent.Both are limiting

reagent.

i) Find the moles of CH4

CH4 : CO2 molar ratio

1 : 1

x : 2

x = 2 mol

Moles of CO2 = 2 mol

ii) Find the moles of O2

O2 : H2O molar ratio

2 : 2

x : 4

x = 4 mol

Moles of O2 = 4 mol

iii) Excess mol of CH4:

Its zero. Because it is a limiting reagent. All the moles of CH4 consumed.

iv) Excess mol of O2:

Its zero. Because it is a limiting reagent. All the moles of O2 consumed in the reaction.

d) Given,

Moles of CO2 = 3 mol

1) Find the moles of CH4

CH4 : CO2 molar ratio

1 : 1

x : 3

x = 3 mol

Moles of CH4 = 3 mol

ii) Find the moles of O2

O2 : CO2

2 : 1

x : 3

x = 6 mol

Moles of O2 = 6 mol

iii) Moles of H2O

H2O : CO2

2 : 1

x : 3

x = 6 mol

Moles of H2O = 6 mol

iii) Excess mol of CH4:

Its zero. Because it is a limiting reagent. All the moles of CH4 consumed.

iv) Excess mol of O2:

Its zero. Because it is a limiting reagent. All the moles of O2 consumed in the reaction.

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Basic stoichiometry Post-lab exercises answers

1) Balance the given equation:

N2 + 2O2 >> 2NO2

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2) Balance the given equation

2SO2 + 3O2 >> 2SO3

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3) Given equation: P4 + 6Cl2 >> 4PCl3

Moles of P4 = 3 mol

Moles of Cl2 = x

P4 : Cl2 molar ratio

1 : 6

3 : x

x = 18 mol

Moles of Cl2 = 18 mol

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4) Given 5 moles of P4 react with 22 moles of Cl2.

The balanced equation is P4 + 6Cl2 >> 4PCl3

a) Moles of PCl3 = x

To find the moles of PCl3, first we have to find out the limiting reagent. To find the limiting reagent, divide by co-efficients of balanced equation:

P4 > 5 / 1 = 5 mol

Cl2 > 22/6 = 3.67 mol

Cl2 has the lower value. So, Cl2 is the limiting reagent. P4 is the excess reagent.

Now find the molar ratio of Cl2 and PCl3

Cl2 : PCl3

6 : 4

22 : x

x = 14.67 mol

Moles of PCl3 = 14.67 mol

b) Find the moles of P4 left in excess after the reaction

P4 : Cl2 molar ratio

1 : 6

x : 22

>> 6x = 22 >> x = 22/6 = 3.67

Moles of P4 reacted = 3.67 mol

moles of P4 left behind = 5 - 3.67 1.33 mol

c) Cl2 is the limiting reagent. So, no more moles of Cl2 left behind after the reaction.

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5) The balanced equation is 2Al + 3Br2 >> 2AlBr3

Given: 3 moles of Al react with4 moles of Br2

To find the limiting and excess reagent, divide Al and Br2 by co-efficients of the balanced equation.

Al >> 3/2 >> 1.5 mol

Br2 >. 4/3 >> 1.33 mol

a) The limiting reagent is Br2. Because it has less moles, Which is consumed completely.

b) The excess reagent is Al. Because it has more moles.

c) Moles of AlBr3 = x

To find the moles of AlBr3, do the molar ratio of Br2 and AlBr3, because Br2 is the limiting reagent.

Br2 : AlBr3

3 : 2

4 : x

3x = 8 >> x = 8/3 = 2.67 mol

Moles of AlBr3 = 2.67 mol

d) Find the moles of excess reagent left behind

Al : Br2 molar ratio

2 : 3

x : 4

3x = 8 > x = 8/3 = 2.67 mol

Moles of Al reacted = 2.67 mol

No of moles of Al left behind = 3 - 2.67 mol = 0.33 mol

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6) Balance the given equation

2Na + Cl2 >> 2NaCl

Given : 4.5 moles of Na reacts with 3.5 moles of Cl2. Find the moles of NaCl.

To find the moles of NaCl, first we have to find the limiting and excess reagent, divide the Na and Cl2 co-efficients of the balanced equation.

Na >> 4.5 / 2 = 2.25 mol

Cl2 >> 3.5 / 1 = 3.5 mol

Na is the limiting reagent, because it has a less moles.

To find the moles of NaCl, we have to do the molar ratio between limiting reagent (Na) and NaCl

Na : NaCl

2 : 2

4.5 : x

2x = 9 >> x = 4.5 mol

Moles of NaCl = 4.5 mol

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