Question

Chemistry help needed!

1. How many grams of hydrogen must you start with to prepare 6.00g of tungsten?(For WO3, MW = 231.8 amu.)

2. What mass of PCl5 will be produced from the given masses of both reactants? PCl5 that can be produced from 22.0g of P4 (and excess Cl2). PCl5 that can be produced from 53.0g of Cl2 (and excess P4).

Thanks!

Answer 1

1)

6 g of hydrogen required for 183.84 g of tungsten

x   g of hydrogen required for 6 g of tungsten

x = 6 x 6 / 183.84= 0.195g of hydrogen is required

2)You need to use
2P + 5Cl2 --> 2PCl5
[or P4 + 10Cl2 --> 4PCl5 if you prefer]

that tells you that 2P = 62g needs 5Cl2 = 355g Cl2 to react

so 22g P needs 355/62 x 22g Cl2 = 125.96g

so you don't have enough chlorine so chlorine is the limiting reagent,.

from eqtn 355g Cl2 gives 479gg PCl5
so 53g Cl2 gives 479/355 x 53 = 70.61g

Answer 2

Balanced equation:
WO₃ + 3 H₂ = W + 3 H₂O
means 6 moles of H₂ gives 1mol of W i.e 6.04764 grms of H₂ gives 183.84 grms of W.
so to prepare 6 grams of W we must start with 0.197 grms of H₂ .

a)

for excess Cl2 , Balanced equation :

P4 + 10Cl2 ------> 4PCl5

22 g P4

determine moles P4
moles = mass / molar mass
= 22.0 g / 123.88 g/mol
= 0.177moles

balanced equation shows that
1 moles P4 reacts to form 4 moles P4Cl5
So 0.177moles P4 can form (4 x 0.177) moles P4Cl5
= 0.71moles P4Cl5 formed

mass = molar mass x moles
mass P4Cl5 = 208.22 g/mol x 0.71mol
= 147.9g of pcl5 formed

b) for excess P4 , Balanced equation :

P4 + 10Cl2 ------> 4PCl5

moles Cl2 = mass / molar mass
= 53.0 g / 70.9 g/mol
= 0.747 moles Cl2 provided

balanced equation shows that
10 moles Cl2 reacts to form 4 moles PCl5
So 1 mole Cl2 gives 4/10 moles PCl5
Thus 0.747 moles Cl2 will give (4/10 x 0.747) moles PCl5
= 0.298 moles PCl5 possible

mass = molar mass x moles
mass PCl5 = 208.22 g/mol x 0.298 moles
= 62.26 g