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Veronika [31]
3 years ago
14

You are performing the below chemical reaction using Reagent A and Reagent B to produce Product C: A + B -> C You are using 2

moles of A and 5 moles of B. How many Equivalents of B are you using?
Chemistry
1 answer:
Luba_88 [7]3 years ago
8 0

<u>Answer:</u> The equivalents of B tat are getting used in the given chemical reaction are 2 moles.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is present in less amount and also it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount in the chemical reaction.

We are given:

Moles of A = 2 moles

Moles of B = 5 moles

For the given chemical reaction:

A+B\rightarrow C

By Stoichiometry of the reaction:

1 mole of A reacts with 1 mole of B

So, 2 moles of A will react with = \frac{1}{1}\times 2=2mol of B

As, given amount of B is more than the required amount. So, it is considered as an excess reagent.

Thus, A is considered as a limiting reagent because it limits the formation of product.

Amount of B used = 2 moles

Hence, the equivalents of B tat are getting used in the given chemical reaction are 2 moles.

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A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the foll
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b) pH = 2.300

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e) pH = 13.736

Explanation:

Both HBr and NaOH are strong acids and bases so they can be considered to be fully dissociated in solution. Therefore the concentration of H+ and OH- can considered to be equal to the concentration of HBr and NaOH respectively.

Number of moles (mol) = Concentration of solution (mol/L) * volume of solution (L)

Step 1: Calculate number of moles of H+ in initial solution

moles of H+ = 0.200 mol/L * 0.02 L = 0.004 mol

Step 2: Calculate number of moles of OH- in titrating solution

a) moles of OH- = 0.200 mol/L * 0.015 L = 0.003 mol

b) moles of OH- = 0.200 mol/L * 0.0199 L = 0.00398 mol

c) moles of OH- = 0.200 mol/L * 0.020 L = 0.004 mol

d) moles of OH- = 0.200 mol/L * 0.0201 L = 0.00402 mol

e) moles of OH- = 0.200 mol/L * 0.035 L = 0.007 mol

Before neutralization point, moles of H+ have to be determined by taking the difference between moles of H+ in initial solution and total moles of OH- added. After neutralization point, moles of OH- have to be determined by taking the difference between total moles of OH- added and moles of H+ in initial solution. pH at neutralization point is 7

Step 3: Calculate moles of H+/OH- remaining

a) moles of H+ = 0.004 - 0.003 = 0.001 mol

b) moles of H+ = 0.004 - 0.00398 = 0.00002 mol

c) moles of H+ = moles of OH- = 0.004 mol (neutralization point)

d) moles of OH- = 0.00402 - 0.004 = 0.00002 mol

e) moles of OH- = 0.007 - 0.004 = 0.003 mol

Total volume of solution has to be determined by adding volume of initial solution and volume of titrating solution added. Concentration of H+/OH- has to be calculated dividing moles of H+/OH- by the total volume on solution.

Step 4: Calculate concentration of H+/OH- after addition of base

a) total volume = 0.002 + 0.0015 = 0.0035 L

[H+] = 0.001 mol / 0.0035 L = 0.286 mol/L

b) total volume = 0.002 + 0.00199 = 0.00399 L

[H+] = 0.00002 mol / 0.00399 L = 0.00501 mol/L

c) total volume = 0.002 + 0.002 = 0.004 L

[H+] = [OH-] = 0.004 mol / 0.004 L = 1 mol/L

d) total volume = 0.002 + 0.00201 = 0.00401 L

[OH-] = 0.00002 mol / 0.00401 L = 0.00499 mol/L

e) total volume = 0.002 + 0.0035 = 0.0055 L

[OH-] = 0.003 mol / 0.0055 L = 0.545 mol/L

The formula to calculate pH from concentration of H+ and OH- is:

pH = -log[H+]

pH = 14 - pOH = 14 + log[OH-]

Step 5: Calculate pH

a) pH = -log(0.286) = 0.544

b) pH = -log(0.00501) = 2.300

c) pH = 7 (neutralization point)

d) pH = 14 + log(0.00499) = 11.698

e) pH = 14 + log(0.545) = 13.736

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