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?

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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2