Question

For the reaction: Al(OH)3 + 3HCl = AlCl3 + 3 H2O
A) How many grams of Al(OH)3 react with 25.00 mL of 0.1038 M HCl solution?
B) What is the molarity of an HCl solution if 71.28 mL of the HCl solution react with 0.1035g of Al(OH)3?
C) What is the limiting reactant when 0.425 g of AL(OH)3 are mixed with 25.0mL of 4.00M HCl solution?
D) How many grams of AlCl3 is produced from the mixture in C?

Answer 1

do you know how to calculate theoretical yield and limiting reactant? do you know the definition of Molarity?

Answer 2

If I follow examples in the text I can sort of work my way through, but not off the top of my head.

Answer 3

the definition of molarity is

M= n/V

n= number of moles of solute
V= volume in liters
first convert the 25.00 mL of 0.1038 M HCl solution to moles of HCl

Answer 4

M= 4.152

Answer 5

no, you have to calculate "n", convert the volume from mL to liters and then multiply the volume (V) times the concentration (M)

Answer 6

.025L* .1038M= .0026

Answer 7

good!
now by the stoiquiometry of the reaction how many moles of Al(OH)3 will react with .0026moles of HCl?

Al(OH)3 + 3HCl = AlCl3 + 3 H2O

Answer 8

.0087

Answer 9

do the calculation again you are out of a decimal place
.0026moles of HCl x 1 mol Al(OH)3 /3 mole HCl = ????? mol Al(OH)3

Answer 10

.00087

Answer 11

now calculate how many grams of Al(OH)3 are .00087 moles of Al(OH)3

number of moles = mass / molecular mass

n=m/MM

Answer 12

mm=78.00
n=.00087
m=.0679?

Answer 13

that is the answer to the question A

Answer 14

for the question B you have to do the opposite process than in A
B) What is the molarity of an HCl solution if 71.28 mL of the HCl solution react with 0.1035g of Al(OH)3?

from g of Al(OH)3 -> moles of Al(OH)3 -> moles of HCl -> Molarity of HCl

Answer 15

.0547M

Answer 16

pretty close I will said is fine

Answer 17

C) What is the limiting reactant when 0.425 g of AL(OH)3 are mixed with 25.0mL of 4.00M HCl solution?

Al(OH)3 + 3HCl = AlCl3 + 3 H2O

0.425 g of Al(OH)3 -> moles of Al(OH)3 -> moles of AlCl3 (1)

V and M HCl -> moles of HCl -> moles of AlCl3 (2)

IF (1) > (2) the limiting reactant is (HCl)
if (1) < (2) the limiting reactant is (Al(OH)3)

Answer 18

Al(OH)3 limiting reactant

Answer 19

D) How many grams of AlCl3 is produced from the mixture in C?
convert the moles of AlCl3 (1) -> grams

Answer 20

.7360g

Answer 21

yes close enough

Answer 22

0.718 g

Answer 23

Thank you