Question

1. How many GRAMS of boron are present in 3.29 moles of boron trifluoride ?

2. How many MOLES of fluorine are present in 2.13 grams of boron trifluoride ?

Please show work, Thanks for the help.

Answer 1

1)

boron trifluoride is BF3

moles of B = moles of BF3 = 3.29 moles

Molar mass of B = 10.81 g/mol

mass of B,

m = number of mol * molar mass

= 3.29 mol * 10.81 g/mol

= 35.6 g

Answer: 35.6 g

2)

Molar mass of BF3,

MM = 1*MM(B) + 3*MM(F)

= 1*10.81 + 3*19.0

= 67.81 g/mol

mass(BF3)= 2.13 g

number of mol of BF3,

n = mass of BF3/molar mass of BF3

=(2.13 g)/(67.81 g/mol)

= 3.141*10^-2 mol

1 BF3 has 3 F

So,

moles of F = 3*number of mol of BF3

= 3* 3.141*10^-2 mol

= 9.42*10^-2 mol

Answer: 9.42*10^-2 mol