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!
