Question

Answer the following question using the data:

A = 85 g/mol

D = 6 g/mol

E = 10 g/mol

G = 74 g/mol

J = 23 g/mol

L = 1 g/mol

M= 36 g/mol

Q = 24 g/mol

R = 87 g/mol

T= 50 g/mol

U = 20g/mol

Z = 44 g/mol

a) A laboratory process for producing A2 is based on the following process:

1. 2E + D2 = 2 ED

2. A2D3 + 2ED = A2 + E2D5

What mass of D2 is required to make 1 x 10^2 kg of A2? show all work

b) Three flasks each contain the same amount of 0.1 mol of GR. Across the mouth of each flask is stretched a deflated balloon. Each balloon has a different amount of T in it: 0.3 g, 1.2 g, 4.8 g. When the T is added to the GR solution, the balloon will fill up with gas based on the following reaction: T (s) + 2 GR (aq)= G2 (g) + TR2 (g)

Fill in the following and show all work.

Flask 1: Moles of T2 _______, Moles of GR _______, Limiting Reagent _________, Moles of G2 produced

Flask 2: Moles of T2 _______, Moles of GR _______, Limiting Reagent _________, Moles of G2 produced

Flask 3: Moles of T2 _______, Moles of GR _______, Limiting Reagent _________, Moles of G2 produced

Answer 1

Points needed to solve :

• To calculate molar mass of a compound, add the atomic masses of each atom from the formula.
• Equation 1 : number of moles = [ (mass in g) / ( molar mass in g/mol) ]
• To find how much substance is needed or produced, it's best to compare the number of moles from the balanced chemical equation.
• Limiting reagant = reagent present in lower quantity than needed for the other reactant to completely react..yield of product is calculated based on amount of limiting reagent taken.

a)

The overall reaction ( sum of the two stepwise reactions) is :

2E + D2 + A2D3 + 2ED = 2ED + A2 + E2D5

Cancelling out 2ED from both sides :

2E + D2 +A2D3 = A2 + E2D5

Molar mass of D2 = 2 (6 g/mol) = 12 g/mol

Molar mass of A2 = 2 ( 85 g/mol) = 170 g/mol.

1x 10^2 kg = 100 kg = 100,000 g of A2 is needed [ 1 kg = 1000 g]

Using equ1, Number of moles of A2 needed = mass /molar mass = 100,000 g / (170 g/mol ) = 588.2 moles

From the balanced equation,

1 mole A2 is formed from 1 mole D2.

So, 588.2 moles A2 is formed from 588.2 moles of D2.

Hence, mass of D2 needed ( use equ1) = number of moles x molar mass = 588.2 mol x 12 g/mol = 7058.4 g of D2 = 7.06 kg of D2 ( answer)

b)

Molar mass of GR = (74 + 87) g/mol = 161 g/mol

Moles of T= mass of T taken / molar mass of T .

Flask 1 : (0.3 g/ 50 g/mol ) = 0.006 moles

Flask 2 : 1.2 g/ 50 g/mol = 0.024 moles

Flask 3 : 4.8 g/ 50 g/mol = 0.096 moles

Moles of GR = 0.1 mol, which is same in all three flasks .

Balanced reaction :

T + 2 GR = G2 + TR2

So, 1 mole of T needs 2 moles of GR to completely react.

Flask 1

0.006 moles of T will need ( 0.006 x 2) = 0.012 moles of GR to completely react.

Amount of GR taken = 0.1 mol = more than needed. So, GR is taken in excess, and there is not enough T to use it up. Hence , T is limiting reagent.

Flask 2

0.024 moles of T will need ( 2 x 0.024) = 0.048 moles of GR.

Again, T is limiting reagent.

Flask 3

0.096 moles of T will need ( 0.096 x 2) = 0.192 moles of GR.

Here, not enough GR is present to fully react with the T. So , T is limiting reagent.

Calculation of moles of G2 produced : this is done based on moles of limiting reagent taken.

Flask 1.

1 mole T gives 1 mole G2.

So, 0.006 moles of T will give 0.006 moles of G2

Flask 2.

0.012 moles of T will give 0.012 moles of G2

Flask 3

2 moles GR gives 1 mole of G2.

So, 0.1 mole of GR will give ( 1/2) x 0.1 = 0.05 moles of G2