Answer :
This is an incomplete question, here is a complete question.
Phosphorus trifluoride is formed from its elements.
[tex]P_4(s)+6F_2(g)\rightarrow 4PF_3(g)[/tex]
How many grams of fluorine are needed to react with 6.20 g of phosphorus?
Answer : The mass of [tex]F_2[/tex] needed are, 11.4 grams.
Explanation : Given,
Mass of [tex]P_4[/tex] = 6.20 g
Molar mass of [tex]P_4[/tex] = 124 g/mol
Molar mass of [tex]F_2[/tex] = 38 g/mol
First we have to calculate the moles of [tex]P_4[/tex]
[tex]\text{Moles of }P_4=\frac{\text{Given mass }P_4}{\text{Molar mass }P_4}[/tex]
[tex]\text{Moles of }P_4=\frac{6.20g}{124g/mol}=0.05mol[/tex]
Now we have to calculate the moles of [tex]F_2[/tex]
The balanced chemical equation is:
[tex]P_4(s)+6F_2(g)\rightarrow 4PF_3(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]P_4[/tex] react with 6 moles of [tex]F_2[/tex]
So, 0.05 moles of [tex]P_4[/tex] react with [tex]0.05\times 6=0.30[/tex] moles of [tex]F_2[/tex]
Now we have to calculate the mass of [tex]F_2[/tex]
[tex]\text{ Mass of }F_2=\text{ Moles of }F_2\times \text{ Molar mass of }F_2[/tex]
[tex]\text{ Mass of }F_2=(0.30moles)\times (38g/mole)=11.4g[/tex]
Therefore, the mass of [tex]F_2[/tex] needed are, 11.4 grams.
The mass of F2 that should be required to react with 6.20 g should be 11.4 grams.
Calculation of the mass:
Since
Mass of P4 is 6.20g
The molar mass of P4 is 124 g/mol.
And, the molar mass of F2 is 38 g/mol.
Now the moles of P4 is
= Given mass / molar mass
= 6.20/124
= 0.05 mol
Now the moels of F2 is
= Moles of P4 * 6
= 0.05 * 6
= 0.30 moles
So, here the mass is
= Moles of F2 * molar mass of F2
= 0.30 * 38
= 11.4 grams
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