If 17.7 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution what is the molar concentration of the ammonia?

Answer 1

0,800 M HCl

= 0,800 moles per 1000 ml (1L)

= 0,0148 moles per 18,5 ml

So you also need 0,0148 moles ammonia to neutralize.

= 0,0148 Moles per 5 ml

= 2,96 Moles per 1000 ml (1L)

So the answer= 2,96 M

Answer 2

If 17.1 mL

m

L

of 0.800 M HCl

M

H

C

l

solution are needed to neutralize 5.00 mL

m

L

of a household ammonia solution, what is the molar concentration of the ammonia?

Answer 3

MV=MV

(17.2mL)(.800M)=M(5.00mL)

2.752 M

2.75 M when you round to the correct amount of significant figures (three)

Answer 4

The molar concentration is 2.91 M of ammonia.

Answer 5

0.0188 L x 0.800 M = 0.01504 mol HCl

0.01504 mol NH4OH / 0.005 L = 3.00 M 3.01 if you are using mastering chemistry.com

Answer 6

2.94 M

.800M=x/.0184L X=.01472 mol

.01472mol/.005 L= 2.94M

Answer 7

The concentration of NH4OH is 2,8 M (or 1,37 NH3).

Answer 8

18.3*.800/5.00=2.92