A student is trying to identify a clear liquid that he finds in a beaker in the classroom. He thinks the liquid might be either water or ethanol, and he decides to identify the liquid using a density. The density of water is 1 g/cm3. The density of ethanol (clear alcohol) is 0.789 g/cm3. The student has 50 cm3 of the liquid, and its mass is 4.99 grams. What is the identity of the liquid?

A. ethanol
B. water
C. neither water nor ethanol
D. a 50/50 mixture of water and ethanol

PLEASE BE RIGHT, WORTH 30 PTS

Simply mulitply the volume by the density. As we shall see, this is dimensionally consistent.

Explanation:

density

ρ

=

Mass

Volume

, and thus units of

g

⋅

m

L

−

1

are reasonable.

For this problem:

17.4

⋅

m

L

×

0.798

⋅

g

⋅

m

L

−

1

≅

14

⋅

g

but A i supposed?

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.