Untitled - Kelly Walsh High School

Mass, Moles, and Equations 47
17. a. Using the limiting reagent from problem 15, we can determine the theoretical
yield: (3.00 mol O2 )a 416 g Br2O3. Inserting the masses into the definition of percent yield gives: Percent yield
2 mol Br2O3 ba
3 mol O2 207.80 mol Br2O3 b
1 mol Br2O3 a 2.40 %
b. 12.2 % c. 2.35 % d. 12.0 % e. 8.67 %
1 mol S
1 mol F
18. (21.4 g S)Q R 0.667 mol S (50.8 g F)Q R 2.67 mol F
32.07 g S 19.0 g F
0.667 mol S/0.667 1 S and 2.67 mol F/0.667 4 F empirical formula is SF4. 1 mol Mn
1 mol O
19. (69.6 g Mn)Q R 1.27 mol Mn (30.4 g O)Q R 1.90 mol O
54.94 g Mn 16.0 g O
1.27 mol Mn/1.27 1 Mn and 1.90 mol O/1.27 1.50, we need to multiply
by two to clear the decimal value (1.50). The empirical formula is Mn2O3. 20.
1 mol S
1 mol F
(25.2 g S)Q R 0.786 mol S (74.8 g F)Q R 3.94 mol F
32.07 g S 19.00 g F
0.786 mol S/0.786 1 S and 3.94 mol F/0.786 5 F, thus, the empirical formula
is SF5. The molar mass of the empirical formula is 127.01 g/mol, which
is half the molar mass given in the problem. This means the empirical formula
is one-half the molecular formula. Therefore, the molecular formula
is S2F10. 10.0 g Br2O3 b 100 %
416gBr2O3