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0.250 g of an element, M, reacts with excess fluorine to produce 0.547 g of the...
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First we have to do the chemical reaction of the unknown element ‘M’ with fluorine.
`M + F_2 -> MF_6`
Balancing it will make it:
`M + 3 F_2 -> MF_6`
We assume that all of the mass of the element M is consumed.
1. Get the mass of F reacted
`Mass MF_6 = mass M + mass F `
`0.547 g = 0.250 + mass F `
`Mass F = 0.297 grams F `
2. get the amount of moles of M via stoichiometry
`0.297 grams F * (1 mol e F)/(19.0 grams F) * (1 mol e M)/(6 mol es F)`
`= 0.002605 mol es M`
3. Identify the unknown element
`At omic mass = (mass of M)/(mol es of M) `
`At omic mass = 0.250/ 0.002605 `
Atomic mass = 95.95 = 96.0 g/mol = Molybdenum (Mo) -> answer
MoF6 -> Molybdenum hexafluoride
Posted by jerichorayel on August 2, 2013 at 11:59 AM (Answer #1)
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