Question

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?

Answer 1

Answer:

$\frac{4molPF_3}{1molP_4}$

Explanation:

Hello

In this case, given the reaction:

$P_4(s)+6F_2(g)\rightarrow 4PF_3(g)$

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

$\frac{4molPF_3}{1molP_4}$

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.

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