[locagata48]

1) What mass of aluminum hydroxide is produced when 50.0 ml of .200 M Al(NO3)3 reacts with 200.0 ml of .100 M KOH?

Al(NO3)3 + 3KOH-> Al(OH)3 + 3KNO3

.05L * .2 M = .01 mol Al(NO3)3
.2L * .1 M = .02 mol KOH

limiting reactant Al(NO3)3

2) Why is liquid volume always read at the bottom of he liquid curve?

Thanks

[Unknown008]

1)Nope for the first question:
1 mol of Al(NO3)3 will react with 3 mol of KOH, OK?
So, proportionally, 0.1 mol of Al(NO3)3 will react with 0.3 mol of KOH, right?
Which one of the two is less than the required amount? You have only 0.2 mol of KOH, instead of 0.3 mol, which you need. Therefore, KOH is limiting.

As a result, you'll have only 2/3 of all of the other reactant participating, that is 2/3 * 0.1 mol of Al(NO3), which is 0.0666 or 1/15. Thus, 1/15 mol of Al(OH)3 will be produced according to the equation, do you agree?

Ok, so mass of Al(OH)3 = Mr of Al(OH)3 * 1/15.

2) As a matter of fact, I was told that it depends on which type of liquid volume I'm measuring. A colourless one (water for example) is easier to read below the meniscus (liquid curve) as this curve is 'practically invisible'.

A dark coloured solution would be measured above the meniscus since this curve can prevent you from reading exactly below the meniscus.

However, in titration, this is not really important if you read above or below the meniscus, provided that if you read above, you should always read above for the other readings, and the same for below the meniscus. But choose the one easier for you to read accurately.

Hoped it helped!