The gram quantity of both reactants is given in the...

The gram quantity of both reactants is given in the following balanced combustion reaction. The mass of CO2 that was (actually) measured at the end of the experiment is also given in parenthesis (in red). (A) Fill all the missing values in the table. (B) Determine which reactant is "limiting." (C) Calculate the percent yield of CO2 for the reaction (D) What is the remaining mass (in g) of the excess reactant? (+2 BONUS points if you get this correct) Again, you can copy and paste (Ctrl+C & then Ctrl+V) this table to the answer box and write your answers there. 2C2H6 + 7O2 ? 4CO2 + 6H2O (30.2 g) Mass, g 15 g 48 g Molar Mass g/mol Moles Mole ratio

Transcript

00:01 Dear students, this is a question on combustion of ethan, the balanced equation is given.

00:06 And in the table, initial mass of ethan and oxygen are given and carbon dioxide form are also given.

00:14 As per the first question, you have to fill all the missing values in the periodic table.

00:18 We have to fill all the missing values.

00:20 Now let us see the first one.

00:23 Molar mass, molar mass of ethan.

00:25 Then molar mass is a carbon atom, so 24 plus 6 equal to 30.

00:35 Molar mass of ethan is 30.

00:38 So this is of ethan and this is of oxygen.

00:43 And molar mass of oxygen is 32.

00:46 Then moles.

00:47 Here we have 2 moles.

00:49 Here we have 7 moles of oxygen.

00:52 Then mole ratio.

00:54 So to calculate the mole ratio, let us first.

00:57 Convert this into moles this is 15 by 30 equal to 0 .5 15 by 30 equal to 0 .5 mole.

01:07 This is ethan and this is 48 by 32, 48 by 32.

01:15 So 16 -3 is 48, 16 -032, so 1 .5.

01:20 So this is 0 .5 and this is 1 .5.

01:24 Now molar ratio, 0 .5 ,000, divided by 2, this is 1 .5 divided by 7.

01:33 So this is approximately 0 .2, this is 0 .05 and this is approximately 0.

01:40 This is 2, then 1 0 means approximately 1.

01:46 So it is 0 .21.

01:49 Now since the oxygen mole ratio is less, therefore oxygen is the limiting reactor.

01:55 So this is limiting reactant.

02:01 Now limiting reactant is oxygen.

02:08 Now coming to the third one, calculate the percent yield of carbon dioxide for the reaction.

02:13 Now since oxygen is the limiting reactant, here we are seeing 7 moles of oxygen gives 4 moles of carbon dioxide.

02:20 Each mole weighs 32...

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