How many millimoles of the substance are present in: 1) 14.2g SnO2 2) 327g FeSO4•C2H4(NH3)2SO4•4H2O

How many millimoles of the substance are present in:

1) 14.2g SnO2

2) 327g FeSO4•C2H4(NH3)2SO4•4H2O

Expert Solution

1)
Masss of SnO2 = 14.2 g
Molar mass of SnO2 = 150.7 g/mol

Number of moles = mass / molar mass
= 14.2 / 150.7
=0.094 mol

Number of millimoles = number of moles * 1000 = 0.094*1000 = 94
Answer: 94 millimoles

2)
Mass = 327 g
Molar mass of FeSO4•C2H4(NH3)2SO4•4H2O = 382.1 g/mol

Number of moles = mass / molar mass
= 327 / 382.1
=0.856 mol

Number of millimoles = number of moles * 1000 = 0.856 *1000 = 856
Answer: 856 millimoles

queen_honey_blossom answered 2 years ago

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