A mixture is made by combining 1.17 lb of salt and 3.54lb of water. What is the percentage of salt (by mass) in this mixture?

The percentage of salt in the mixture 24.84%

Explanation:

In this problem we are required to solve for the percentage of salt in the mixture of salt and water

Step one :

We can proceed by first solving for the total mass of the mixture

Which is given as

mass of mixture =mass of salt +mass of water

Mass of mixture = 1.17+3.54

Mass of mixture = 4.71 lb

Step two :

Now let us calculate the percentage of salt present in the mixture which is

=(mass of salt/mass of mixture) *100

=(1.17/4.71)*100

=0.248*100

=24.84%

petroleum? coal maybe?

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.