Question

Consider the reaction 2 Pb(s) + O2(g) → 2 P50(s) An excess of oxygen reacts with 451.4 g of lead, forming 344.8 g of lead(II)oxide. Calculate the percent yield of the reaction percent yield:

Answer 1

Answer:

Step 1: Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 2: write the balanced chemical equation.

2 Pb(s) + O2(g) --------> 2 PbO(s)

Step 3: calculate the moles of lead (Pb)

we know, moles = mass given / molar mass

mass of lead given = 451.4 g

molar mass of lead  = 207.2 g/mol

moles of ethanol = ( 451.4 g / 207.2 g/mol ) = 2.1786 mol

Step 4: Calculate the moles of lead(ii) oxide

2 Pb(s) + O2(g) --------> 2 PbO(s)

According to the reaction:
2 mol Pb of produce 2 mol of PbO
so, 2.1786 mol of Pb will produce = 2.1786 mol of PbO

Step 5: Calculation of mass of lead(ii) oxide (PbO) produced

we get moles of lead(ii) oxide  that can be produced = 2.1786 mol

Molar mass of lead(ii) oxide = 223.2 g/mol

Mass of lead(ii) oxide  produced =( moles×molar mass ) = ( 2.1786 mol × 223.2 g/mol ) = 486.26 g

step 6: Calculation of percent yield of lead(ii) oxide

Given,

actual yield = 344.8 g

We got theoretical yield = 486.26 g

We know, Percent yield = (actual yield / theoretical yield ) × 100 %

Percent yield = ( 344.8 / 486.26 ) × 100 % = 70.9 %

Hence, the percent yield of reaction = 70.9 %