Answer to Question #161375 in General Chemistry for Tyler Murphy

Question

1. You can perform a variation of the zinc and sulfur reaction we discussed yesterday, by reacting sulfur with powdered lithium instead, as shown below. If you perform the reaction with 15.0 g of lithium and 15.0 g of sulfur:

16 Li(s)  +   S8(s)  →  8 Li2S(s)

1. What is the limiting reactant? {/10}
2. What mass of product can you make? {/4}
3. What mass of excess reactant will be leftover? {/6}

Answer 1

15g Li × $\frac{1moleLi}{6.941gLi}$ = 2.161 moles Li

15g S₈ × $\frac{1moleS_8}{256.52gS_8}$ = 0.058 moles S₈

1). there are 2.161 moles of Li and 0.058 moles of S₈

To use all the Li , how many moles of S₈ do we need

2.161 mole Li × $\frac{1moleS_8}{16moleLi}$ = 0.135 mole S₈

To use all the S₈ , how many moles of Li do we need

0.058 moles S₈ × $\frac{16moleLi}{1moleS_8}$ = 0.928 mole Li

We have enough Li to use all the S₈ but we dont have enough S₈ to use all the Li .

S₈ is the limiting reactant .

2). How many moles of Li₂S can we use all the S₈

0.058 mole S₈ × $\frac{8moleLi_2S}{1moleS_8}$ = 0.464 moles

How many grams of Li₂S is this

= 0.0464 × 45.95

= 21.3208 g

21.3208 g mass of the product we made

3). Al is the excess reactant

how much are left over

Excess reactant = Total reactant - used reactant

Excess reactant = 2.161 mole - 0.928 mole

= 1.233 moles of Li are left

1.233 mole Li × $\frac{6.941g Li}{1moleLi}$ = 8.558 g

8.558 g of Li are left over