Question

How many moles of Mn(CO₃)₂•2H₂O are present in 727.251 g of the compound? Report your answer to 2 decimal places. Do not round for the whole question, use every number until you get the final answer.

Answer 1

Molar mass of Mn(CO3)2•2H2O = 1*MM(Mn) + 2*MM(C) + 8*MM(O) + 4*MM(H)

= 1*54.94 + 2*12.01 + 8*16.0 + 4*1.008

= 210.992 g/mol

mass of Mn(CO3)2•2H2O = 727.251 g

we have below equation to be used:

number of mol of Mn(CO3)2•2H2O,

n = mass of Mn(CO3)2•2H2O/molar mass of Mn(CO3)2•2H2O

=(727.251 g)/(210.992 g/mol)

= 3.45 mol

Answer: 3.45 mol