When the ore lead(II) sulfide burns in oxygen, the products are solid lead(II) oxide and sulfur dioxide gas.a. Write the balanced equation for the reaction.b. How many grams of oxygen are required to react with 29.9 g of lead(II) sulfide?c. How many grams of sulfur dioxide can be produced when 65.0 g of lead(II) sulfide reacts?d. How many grams of lead(II) sulfide are used to produce 128 g of lead(II) oxide?
Solution 57 QPHere, we are calculating the each of the following.Step 1 of 4a.From the given,Lead (II) sulfide burns in the presence of Oxygen and to form lead oxide and sulphur dioxide.A chemical reaction is as follows.PbS(s) + O2(g) PbO(s) + SO2(g).Let’s balance the above chemical reaction Atoms Reactants ProductsBalanced/not balanced Pb 1 1Balanced S 1 1Balanced O 2 3Not balancedTo balance the equation, Subscript of O2 is 2 it is used as a coefficient for PbS ,PbO and SO2.And place the coefficient 3 in front of O2.2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g). Atoms Reactants ProductsBalanced/not balanced Pb 2 2Balanced S 2 2Balanced O 6 6 BalancedTherefore, the balanced chemical reaction in between lead(II) sulfide and oxygen is given below.2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g).____________________________________________________________________________