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What is the mole ratio needed to determine the mass of phosphorus trifluoride produced from the reaction of 120 g of phosphorus with excess fluorine?

What Is The Mole Ratio Needed To Determine The Mass Of Phosphorus Trifluoride Produced From The Reaction Of 120 G Of Phosphorus With Excess Fluorine class=

Sagot :

Answer:

[tex]\frac{4molPF_3}{1molP_4}[/tex]

Explanation:

Hello

In this case, given the reaction:

[tex]P_4(s)+6F_2(g)\rightarrow 4PF_3(g)[/tex]

It means that since the coefficients preceding phosphorous and phosphorous trifluoride are 1 and 4, the correct mole ratio should be:

[tex]\frac{4molPF_3}{1molP_4}[/tex]

Because given the mass of phosphorous it is convenient to convert it to moles and then cancel it out with the moles on bottom of the mole ratio.

Bes regards!

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