The complete combustion of ethene (C2H4) produces water and carbon dioxide. What will occur for this reaction if 75.8 g of ethene reacts with 201 g of O2? a. Identify the limiting reagent. b. Calculate the mass of water produced

A. O2 is the limiting reactant.

B. 75.38g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

C2H4 + 3O2 —> 2CO2 + 2H2O

Next, we shall determine the masses of C2H4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of C2H4 = (12x2) + (4x1) = 28g/mol

Mass of C2H4 from the balanced equation = 1 x 28 = 28g

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

28g of C2H4 reacted with 96g of O2 to produce 36g of H2O.

A. Determination of the limiting reactant. This can be achieved as shown below:

From the balanced equation above,

28g of C2H4 reacted with 96g of O2.

Therefore, 75.8g of C2H4 will react with = (75.8 x 96)/28 = 259.89g of O2.

From the illustration above, we can see that a higher mass i.e 259.89g of O2 than what was given i.e 201g is needed to react completely with 75.8g of C2H4.

Therefore, O2 is the limiting reactant and C2H4 is the excess reactant.

B. Determination of the mass of water, H2O produced from the reaction. This is illustrated below:

In this case the limiting reactant is used because it will give the maximum yield of water as all of it is used up in the reaction. The limiting reactant is O2.

From the balanced equation above,

96g of O2 reacted to produce 36g of H2O.

Therefore, 201g of O2 will react to produce = (201 x 36)/96 = 75.38g of H2O.

Therefore, 75.38g of H2O were produced from the reaction.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"