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HELP! PLEASE! ASAP! Manganese(III) fluoride, MnF3, can be prepared by the following reaction:
2MnI2(s) + 13F2(g) → 2MnF3(s) + 41F5(l)
If the percentage yield of MnF3 is always approximately 57.2%, how many grams of MnFz can be expected if 55.0 grams of each reactant is used in an
experiment?​

Respuesta :

cristisp93

Answer:

14.336 g MnF₂

Explanation:

number of moles = mass / molecular weight

number of moles of MnI₂ = 55 / 309 = 0.178 moles

number of moles of F₂ = 55 / 38 = 1.447 moles

From the reaction and the number of moles calculated we deduce that the fluorine F₂ is a limiting reactant.

So:

if        13 moles of F₂ reacts to produce 2 moles of MnF₃

then   1.447 moles of F₂ reacts to produce X moles of MnF₃

X = (1.447 × 2) / 13 = 0.223 moles of MnF₃ (100% yield)

For 57.2% yield we have:

number of moles of MnF₃ = (57.2 / 100) × 0.223 = 0.128 moles

mass = number of moles × molecular weight

mass of MnF₃ = 0.128 × 112 = 14.336 g

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