Question

Excess solid magnesium is added to 2.50 moles of aqueous lead (IV) nitrate. How many grams of dried solid product should be formed?

Answer 1

518 grams

Explanations:

The reaction between magnesium and lead(IV)nitrate is given as:

`2Mg+Pb(NO_3)_4\rightarrow Pb+2Mg(NO_3)_2`

Given the following parameters

moles of aqueous lead (IV) nitrate = 2.50moles

According to stoichiometry, 1 mole of lead(IV)nitrate produce 1 mole of solid lead, the moles of lead required will be 2.50moles

Determine the mass of solid lead product

Mass of Pb = moles * molar mass

mass of Pb = 2.5 * 207.2

Mass of Pb = 518grams

Hence the mass of dried solid product that should be formed is 518 grams