Question

Part B How many moles of CO are produced when 2.2 moles C reacts? ication Express your answer using two significant figures. к" ANSWER pr moles Part C How many moles of SO2 are required to produce 0.55 mole CS2? Express your answer using two significant figures. ANSWER moles Part D How many moles of CS2 are produced when 3.0 moles C reacts? Express your answer using two significant figures. ANSWER moles Problem 7.76 - Enhanced - with Feedback You may want to reference (Pages 253-258) Section 7.9 while completing this problem. For each of the following reactions, 25.0 g of each reactant is present initially Part A 4Al(s) +302(g)- 2A1203(s) Given 25.0 g of Al, how many grams of the product Al2 O3 could be produced? Express your answer with one decimal place and with appropriate units. ANSWER Part B 4Al(s)+302 (g)2Al2Os (s) Given 25.0 g of O2, how many grams of the product Al2 Os could be produced?

Answer 1

Ans 7.76 :

Part A :

Mol Al = mass / molar mass

= 25.0 g / 26.98 g/mol = 0.926 mol

4 mol Al makes 2 mol Al2O3

so 0.926 mol Al will form : 0.926 mol /2 = 0.463 mol Al2O3

then mass of Al2O3 produced = mol x molar mass

= 0.463 mol x 101.96 g/mol

= 47.2 g

Part B :

Mol O2 = mass / molar mass

= 25.0 g / 31.999 g/mol = 0.781 mol

3 mol O2 needs 2 mol Al

so 0.781 mol O2 will need : (0.781 mol x 2 mol) / 3 mol = 1.04 mol Al

but here only 0.926 mol Al is present , so Al is limiting reagent

so here again mass of Al2O3 produced will be 47.2 g

Part E : NO₂

Mol NO2 = mass / molar mass

= 25.0 g / 46.0055 g/mol = 0.543 mol

Mol H2O = 25.0 g / 18.01528 g/mol = 1.39 mol

each mol H2O needs 3 mol NO2 , here NO2 present is less , so limiting reactant here is NO₂

Part F :

3 mol NO2 produces 2 mol HNO3

so 0.543 mol NO2 will produce : (0.543 mol x 2 mol) / 3 mol

= 0.362 mol HNO3

mass of HNO3 = mol x molar mass

= 0.362 mol x 63.01 g/mol

= 22.8 g

Part G : O2

Mol C2H6O = mass / molar mass

= 25.0 g / 46.07 g/mol = 0.543 mol

Mol O2 = 25.0 g / 31.999 g/mol = 0.781 mol

each mol C2H6O needs 3 mol O2

so 0.543 mol C2H6O will need : 0.543 mol x 3 = 1.6 mol O2

here only 0.781 mol O2 is present , so limiting reactant is : O2