Question

When lead(ii) nitrate is heated it decomposes according to the reaction 2Pb(NO3)2 ==> 2PbO +4NO2 + 02. Calculate the mass of lead(ii) nitrate needed to produce 5.0g of oxygen.

Answer 1

Molar mass of:

$Pb(NO_3)_2=331g\\O_2=32g$

2 moles of Lead (ii) Nitrate yields 1 Mole of Oxygen gas.

Therefore $(2×331)g \>Pb(NO_3)_2\>$

produce 32g of Oxygen gas.

5g of Oxygen will be produced by:

$\frac{5×2×331}{32}=103.4375g$