₦airaland Forum

A student conducting this experiment weighs out 3.10g of CoCl2.6H2O.
For the part you will be calculating the amount Of sodium carbonate required for the complete precipitation of cobalt2 carbonate.
Calculate the moles of CoCl2.6H2O used in this reaction. Thanks guys.

Are you asking how to calculate just the moles??

If yes, the use the formula

Moles = Mass/Molar mass of CoCl2.6H2O

Let MM = Molar Mass

MM of CoCl2.6H2O = MM of Co + (2*MM of Cl) + ( 6*2 MM of H) + (6*MM Of O)

From the periodic tables,

MM of Co = 58.93 g/mol

MM of Cl = 35.45 g/mol

MM of H = 1.00 g/mol

MM of O = 16.00g/mol

MM of CoCl2.6H2O = 58.93 + (2*35.45)+(12*1)+(6*16)

MM of CoCl2.6H2O = 237.83 g/mol

Moles of CoCl2.6H2O = 3.10g/237.83g/mol

Moles of CoCl2.6H2O = 0.013 moles

raayah:
Are you asking how to calculate just the moles??

If yes, the use the formula

Moles = Mass/Molar mass of CoCl2.6H2O

Let MM = Molar Mass

MM of CoCl2.6H2O = MM of Co + (2*MM of Cl) + ( 6*2 MM of H) + (6*MM Of O)

From the periodic tables,

MM of Co = 58.93 g/mol

MM of Cl = 35.45 g/mol

MM of H = 1.00 g/mol

MM of O = 16.00g/mol

I love you bro. No homo. God bless you.

cazchukwu:
I love you bro. No homo. God bless you.

Ok,I have updated the full calculation,if this is what you're looking for

raayah:
Ok,I have updated the full calculation,if this is what you're looking for

Just noticed that you're a female. Thanks tho. The answer was right. Passed my Online quiz cos of it.