Calculate the mass percent of oxygen in chromium(iii) carbonate. round your answer to the nearest percentage.

The percentage is 50.7%

Explanation:

The formula of chromium (III) carbonate is (Cr₂(CO₃)₃)

Let us calculate the molar mass of (Cr₂(CO₃)₃)

molar mass of (Cr₂(CO₃)₃) =

2X atomic mass of Cr + 3X atomic mass of C + 9X atomic mass of O

molar mass of (Cr₂(CO₃)₃) = 2X 52 + 3X12 +9X16 = 284g/mol

As per the formula there are nine moles of oxygen in each mole of (Cr₂(CO₃)₃).

so we can say that

In 284g of (Cr₂(CO₃)₃) the mass of oxygen = 144g

so 1 g of (Cr₂(CO₃)₃) the mass of oxygen =

Therefore

the mass of oxygen in 100g of(Cr₂(CO₃)₃) =0.507X100=50.7

Thus percentage of Oxygen in (Cr₂(CO₃)₃) =50.7%

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.

56 moles of   lieter in chemistry