1.

How many moles of P_(4) can be produced by reaction of 0.10 moles Ca_(5)(PO_4)_(3) F, 0.36 moles SiO_(2) and 0.90 moles C according to the following reaction? 4Ca_(5)(PO_4)_(3)F + 18SiO_(2) + 30 Cto 3P_(4) + 2CaF_(2) + 18CaSiO_(3)+30CO.

Answer»

<P>`0.060`
`0.030`
`0.045`
`0.075`

SOLUTION :`SiO_(2)` = Limiting REACTANT
`18 SiO_(2) overset("give")(rarr) 3P_(4)`
`0.36 mol SiO_(2) to 3/18 xx 0.36 = 0.06 mol P_(4)`.


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