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strojnjashka [21]
3 years ago
7

How many moles of silver oxide (l) are needed to produce 4 moles of silver?

Chemistry
1 answer:
ozzi3 years ago
7 0

The number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles

<h3>Stoichiometry </h3>

From the question, we are to determine the number of moles of silver oxide (I) needed to produce 4 moles of silver

First, we will write the balaced chemical equation for the decomposition of silver oxide (I)

2Ag₂O(s) → 4Ag(s) + O₂(g)

This means, 2 moles of silver oxide (I) [Ag₂O] decomposes to give 4 moles of <u>silver </u>and 1 mole of oxygen gas.

From the <em>balanced chemical equation</em>, it is easy to deduce the number of moles of silver oxide (I) that would give 4 moles of silver.

Hence, the number of moles of silver oxide (I) needed to produce 4 moles of silver is 2 moles

Learn more on Stoichiometry here: brainly.com/question/18834543

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The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH):
Nadya [2.5K]

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

       

6 0
3 years ago
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