Question

2 A(s) + B ------> C + 2 D

Use 44 grams/mole as the molar mass of D and 36 grams/mole as the molar mass of A and 58 grams/mole as the molar mass of B.

If 27 mL of 6.0 M B is needed to react with all of the A, how many grams of D is formed? What mass of A reacted?

What is the Molarity of A if 21.4 mL of 0.355M B is required to neutralize 25.3 mL of A?

What Volume of A whose concentration is 0.745 M is needed to neutralize 23.4 mL of 0.545 M B?

Calculate the percent of B in a sample given that 0.789 grams of that sample required 24.5 mL of 0.275 M of A for neutralization.

PLEASE SOLVE EACH PART OF THE PROBLEM AND SHOW ANYWORK INVOLVED BEFORE COMING TO THE ANSWER

Answer 1

V = 27 ml of M = 0.6 B

mol of B used

mol = MV = 27*0.6 = 16.2 mmol of B..

ratio is 1:2

so

16.2*2 = 32.4 mmol of D ar eproduced

mass = mol*MW = (32.4*10^-3)(44) = 1.4256 g of D

b)

mass of A reacted

mol of A = 2*mol of B

mol of B = 16.2 mmol

mmol of A = 2*16.2 = 32.4 mmol of A

mass = mol*MW = 32.4*10^-3)(36) = 1.1664 g of A

c)

molarity of A if

M of A = mmol/ml = 32.4/25.3 = 1.2806 M

d)

% of B in sample

m = 0.789 g of sample

requires

mol of A = 0.275*24.5 = 6.7375mmol of A

then

mmol of B = 1/2*A = 1/2*6.7375 = 3.36875 mmol of B

mass = mol*MW = (3.36875*10^-3)(58) = 0.1953875 g of B

then

% mass = mas sof B / total mass * 100 = 0.1953875/(0.789 )*100 = 24.763% of B