You need 270 ml of a 65% alcohol solution. on hand, you have a 90% alcohol mixture. how much of the 90% alcohol mixture and pure water will you need to obtain the desired solution?

You must add 75 mL water to 195 mL 90 % alcohol to make 270 mL of 65 % alcohol.

Step 1. Calculate the volume of 90 % alcohol needed

You can use the dilution formula

V1×C1 = V2×C2

where

V1 and V2 are the volumes of the two solutions

C1 and C2 are the concentrations

You can solve the above formula to get

V2 = V1 × C1/C2

V1 = 270 mL; C1 = 65 %

V2 = ?; C2 = 90 %

∴V2 = 270 mL × (65 %/90 %) = 195 mL

You need 195 mL of 90 % alcohol to make 270 mL of 65 % RA

Step 2. Calculate the amount of water to add.

Volume of water = 270 mL – 195 mL = 75 mL

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.