Question

n 5 of 10 > Consider the reaction. 2 Pb(s) + 0,(g) 2 PbO(s) An excess of oxygen reacts with 451.4 g of lead, forming 363.3 g of lead(II)oxide. Calculate the percent yield of the reaction percent yield: 173.4

Answer 1

2 moles of lead gives 2 moles of lead(II) oxide.

1 mole of Pb will give 1 mole PbO

Number of moles = mass/molar mass

Molar mass of Pb =207.2 g/mol

Moles of Pb = 451.4/207.2

= 2.18 moles Pb

Moles of PbO formed = 2.18 moles

Molar mass of PbO = 223.2 g/mol

Mass of PbO formed = 223.2 g/mol × 2.18 mol

= 486.58 g

Percent yield = (actual yield/theoretical yield) × 100

Percent yield =( 363.3/486.58)×100

= 74.66 %