Lead ion can be precipitated from solution with nacl according to the reaction: pb2 (aq)+2nacl(aq) pbcl2(s)+ 2na (aq) when 135.8 g of naci are added to a solution containing 195.7g of pb2*, a pbcl2 precipitate forms. the precipice is filtered and dried ad found to have a mass of 252.4g. determine the limiting reactant, theoretical yield of pbc12, and percent yield for the reaction

is limiting reagent.

Theoretical yield = 262.67 g

% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %

Explanation:

The formula for the calculation of moles is shown below:

Given: For

Given mass = 135.8 g

Molar mass of = 58.44 g/mol

Moles of = 135.8 g / 58.44 g/mol = 2.3238 moles

Given: For

Given mass = 195.7 g

Molar mass of = 207.2 g/mol

Moles of = 195.7 g / 207.2 g/mol = 0.9445 moles

According to the given reaction:

1 mole of react with 2 moles of

Also,

0.9445 mole of react with 2*0.9445 moles of

Moles of = 1.889 moles

Available moles of = 2.3238 moles  (Extra)

Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of gives 1 mole of

0.9445 mole of gives 0.9445 mole of

Mole of  = 0.9445 moles

Molar mass of = 278.1 g/mol

Mass of = Moles × Molar mass = 0.9445 × 278.1 g = 262.67 g

Theoretical yield = 262.67 g

Given experimental yield = 252.4 g

% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm