Answer:
The answer to your question is:
a) BaCl2
b) 0.8208 g
c) yield = 85.3 %
d)
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
