Question

The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):

BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)
If you have to prepare the reactants by dissolving 1.00 g of BaCl₂ and 1.00 g of NaOH in water,
(a) What is the limiting reactant?
(b) How many grams of Ba(OH)₂ are produced?
(c) If your experiment produced 0.700 g of Ba(OH)₂, what is the percent yield of Ba(OH)₂?
(d) Based on this percent yield, how much limiting reactant should be used to achieve the target Ba(OH)₂ theoretical yield?

Answer 1

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

d)

Step-by-step explanation:

BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

208 g of BaCl2 ------------- 1 mol

1 g of BaCl2 ------------- x

x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

40 g of NaOH ------------ 1 mol

1 g of NaOH ------------ x

x = (1 x 1) / 40

x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

b)

1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

0.0048 mol --------------- x

x = (0.0048 x 1) / 1

x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

171 g of Ba(OH)2 -------------------- 1 mol

x -------------------- 0.0048 mol

x = (0.0048 x 171) / 1

x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

% yield = 0.700 / 0.8208 x 100

% yield = 85.3

d)

Sorry, i don't understand this question