Moles question (1 Viewer)

[mtsmahia]

Hi, can anyone help solve this questions

Q) A compound of mass 2.41g is obtained when 1.77g of coblt is reacted with oxygen. The emperical formula of the compund is ?



and

Q) Excess dilute sulfuric acid was added to 3.00g of Calcium Carbonate to form Calcium sulfate, carbon dioxide and water. Calculate the masses of:

A) sulfuric acid consumed
B) carbon dioxide produced
C)Calcium sulfate formed.

and lol..i know its alot.... but i was wandering how do u recognise from the formula, if the forumla is a molecule?

thanks!!

 

[shady145]

wats the valency of cobalt as it is a transition metal

 

[shady145]

for the second part write a balanced chemical equation
H2SO4 + CaCO3 --------- > CaSO4 + CO2 + H2O

so they combine in ratios of 1:1 therefore the mole ratios are 1:1.
now work out the moles for calcium carbonate
n=m/mm=3/100.09=0.03
so thats means, becasue the mole ratio is 1:1 then .03 mole of sulfuric acid is consumed, .03 mole carbon dioxide is produced, and 0.3 mole calcium sulphate. u can work out using n=m/mm to find massess of each needed

 

[mtsmahia]

^ thats what i did and i got the wrong answer, well that probably measn the anwers are wrong.... what about the first Q? any ideas

 

[Fortify]

Co = 1.77g

O = 2.41g - 1.77g = 0.64g

n(Co) = m / mm = 1.77/58.93 = 0.030035635

n(Co) = 0.03

n(O) = 0.64/16 = 0.04

n(Co) = 0.04

Ratio is 0.03 : 0.04 = 3 : 4

Therefore Empirical Formula is Co3O4.

 

[Fortify]

CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O

Ratio is 1 : 1 : 1 : 1 : 1

n(CaCO3) = m / mm = 3.00 / (40.08+14.01+16+16+16) = 3.00 / 102.9 = 0.029385836

N(CaCO3) = 0.03mol

Since the molar ratio is 1 : 1 : 1 : 1 : 1, all the will have the same moles.

Therefore:
n(CaCO3) = n(H2SO4) = n(CaSO4) = n(CO2) = n(H2O)

A)
n(H2SO4) = m / mm

m(H2SO4) = n(H2SO4) * mm = 0.03 * 98.086 = 2.94258g

m(H2SO4) = 2.94g

B)
n(CO2) = m / mm

m(CO2) = n(CO2) * mm = 0.03 * 44.01 =1.3203g

m(CO2) = 1.32g

C)
n(CaSO4) = m / mm

m(CaSO4) = n(CaSO4) * mm = 0.03 * 136.51 = 4.0953G

m(CaSO4) = 4.10g

 

[mtsmahia]

Co = 1.77g

O = 2.41g - 1.77g = 0.64g

n(Co) = m / mm = 1.77/58.93 = 0.030035635

n(Co) = 0.03

n(O) = 0.64/16 = 0.04

n(Co) = 0.04

Ratio is 0.03 : 0.04 = 3 : 4

Therefore Empirical Formula is Co3O4.

how did u get 58.93 as the molar mass of Cobalt, shoudnt it be 12.01+ 16?

 

[sle3pe3bumz]

how did u get 58.93 as the molar mass of Cobalt, shoudnt it be 12.01+ 16?

ahahahahahahahahahahha.

Sorry. I just found that hilarious.

Cobalt isnt carbon & oxygen. It is an element itself. Look at your periodic table. Atomic number 27.

 

[MaNiElla]

hehehehehehe

 

[Fortify]

ahahahahahahahahahahha.

Sorry. I just found that hilarious.

Cobalt isnt carbon & oxygen. It is an element itself. Look at your periodic table. Atomic number 27.

Yep ! Hence Co not CO

 

[mtsmahia]

ahahahahahahahahahahha.

Sorry. I just found that hilarious.

Cobalt isnt carbon & oxygen. It is an element itself. Look at your periodic table. Atomic number 27.

OMG>... kicking myslef for that!!! lol...thanks