Question

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PCl5 can be produced by the reaction of phosphorus trichloride with excess chlorine as follows: PCl3 + Cl2 → PCl5

If the actual yield is 127 g of PCl5 and the percent yield is 84.8%, what mass of PCl3 must be used? Hint: Use actual yield and percent yield to find theoretical yield and proceed from there.

a. 150 g

b. 98.8 g

c.108 g

d.83.8 g

e.71.1 g

2. If the reaction N₂ + 3 H₂ → 2 NH₃ is carried out using 1.40 g of N₂ and 0.400 g of H₂, what mass of excess reactant will remain?

a. 0.200 g N₂

b. 1.00 g N₂

c. Both reactants will be completely consumed

d. 0.100 g H₂

e.0.300 g H₂

_3. How many moles of bromine will react with 0.0500 mole of C₂H₂ in the reaction C₂H₄ + Br₂ → C₂H₄Br₂?

a. 0.186 mol

b. 2.00 mol

c. 0.0250 mol

d. 0.100 mol

e. 0.0500 mol

Answer 1

Percentage yield =  

actual yield Thcoreicaluied100 x 100 Theoreticalyield

  

Theoretical yield-actualyield percentageyield × 100

1279 × 100 Theoretical yield

Theoretical yield- 149.76q

According to the equation PCl₃ + Cl₂$\rightarrow$ PCl₅

One mole of PCl₃ should produce one mole of PCl₅ when theoretical yeild is calculated it is assumed that the number of moles PCl₅ will be formed will be equal to the number of moles of the limiting reactant ( reactant present in lesser concentration )

Therefore number of moles of PCl₅ formed theoretically = number of moles of PCl₃

Molar mass of PCl₅ = 208.24 g / mol

theoretically number of moles of PCl₅ =  

theoretical yieldo f PCl5 molarmasso PC

=  

149.76 208.24

= 0.719 moles

Therefore number of moles of PCl₃ = 0.719 moles

molar mass of PCl₃ = 137.33 g / mol

mass of PCl₃ =  ( number of moles of PCl₃ ) $\times$ ( molar mass of PCl₃ )

= 0.719 moles $\times$ 137.33 g / mol

= 98.74 g

= 98.8 g ( b )

2)

According to the equation N₂ + 3 H₂$\rightarrow$ 2 NH₃

one mole of N₂ reacts with 3 moles of H₂

mass of N₂ = 1.40 g

molar mass of N₂ = 14 g / mol

number of moles of N₂ = \

massofN2 molarmasso f N2

= $\frac{1.4 }{14 }$

= 0.1 moles

mass of H₂ = 0.40 g

molar mass of H₂ = 2 g / mol

number of moles of H₂ =  

masso fH2 molarmasso f H2

=

4

= 0.2 moles

For N₂ to completely react There should be 3 $\times$ 0.1 = 0.3 moles of H₂ since number of moles of H₂  = 0.2 moles < 0.3 H₂ will be the limiting reagent and N₂ will be left

number of moles of N₂ will be left = number of moles of N₂ -

numberofmoleso f H2

= 0.1 -

02

= 0.1 - 0.067

= 0.033 moles

mass of N₂ will be left = number of moles of N₂ will be left $\times$ molar mass of N₂

= 0.033 $\times$ 14

= 0.462 g

3)

C₂H₄ + Br₂$\rightarrow$ C₂H₄Br₂

According to the equation one mole of C₂H₄ reacts with one mole of Br₂

Therefore 0.0500 moles of C₂H₄ reacts with 0.0500 moles of Br₂