An industrial vat contains 650 grams of solid lead (II) chloride formed from a reaction of 870 grams of lead (II) nitrate with excess hydrochloric acid. This is the equation of the reaction: 2HCl + Pb (NO3) 2 → 2HNO3 + PbCl2. What is the percent yield of lead (II) chloride? The percent yield of lead chloride is %

89%

Explanation:

Round it up

88.98 %

Solution:

The Balance Chemical Equation is as follow,

2 HCl + Pb (NO₃) ₂ → 2 HNO₃ + PbCl₂

According to equation,

331.2 g (1 mole) Pb (NO₃) ₂ produces = 278.1 g (1 mole) PbCl₂

So,

870 g of Pb (NO₃) ₂ will produce = X g of PbCl₂

Solving for X,

X = (870 g * 278.1 g) : 331.2 g

X = 730.5 g of PbCl₂

Therefore,

Theoretical Yield = 730.5 g

Also as given,

Actual Yield = 650 g

So using following formula for percentage yield,

%age Yield = (Actual Yield / Theoretical Yield) * 100

Putting values,

%age Yield = (650 g / 730.5 g) * 100

%age Yield = 88.98 %