The next few questions will all relate to the reaction of...

The next few questions will all relate to the reaction of BaCl2 and AgNO3 to produce AgCl and Ba(NO3)2. a. Write and balance the chemical equation. b. Classify the type of reaction (single displacement, double displacement, combustion, synthesis, or decomposition) c. If 25.0 mL of 0.218 M BaCl2 and 30.0 mL of 1.3 M AgNO3 are reacted, what is the theoretical yield of AgCl? Change to concentrations d. If 1.22 g of AgCl are produced, what is the % yield? e. What is the final concentration of the Ba(NO3)2 that is also produced

Transcript

00:01 Alright, so we're working with the reaction between barium chloride and silver nitrate.

00:05 We're told the equation, we just have to write it out and balance it.

00:09 So, barium chloride is bacl2, it's reacting with silver nitrate, which is agno3.

00:19 And the products, it's a double replacement reaction, we're told that the products are silver chloride, agcl, and barium nitrate, bano32.

00:26 Two.

00:29 Okay, so to balance this chemical equation, which is part a, we have two chlorides and two, so two chlorides here, so let's put a two out in front, then that makes two silvers, so we'll put a two out in front, that balances the nitrates as well.

00:45 Okay, what type of reaction is it? well, i already gave that away when i was sort of introducing it.

00:51 It's a double replacement when you have two ionic compounds that are reacting together, and you see that the products are the result of interchanging either the two cations or the two anions.

01:02 So like silver is with nitrate here, it becomes with chloride in the products, and barium becomes with nitrate, so you've exchanged the ions, it's double replacement.

01:12 Alright, then for part c.

01:16 We have 25 ml of some barium chloride and 30 ml of some concentration of silver nitrate.

01:23 What is the theoretical yield of silver chloride? so it's a limiting reagent problem.

01:29 We start with 25 mls of our barium chloride.

01:34 The concentration is 0 .218 molar, so that's 0 .218 moles of bacl2 in every 1000 ml by definition of molarity, writing it as ml so the units are consistent here.

01:47 Then we use the mole ratio.

01:49 We're looking for agcl, so that's a 2.

01:51 2 moles of agcl will form for every 1 mole bacl2.

02:02 And let's see.

02:03 We're reacting 25 milliliters with 30 milliliters of the other thing, so the total volume would be 55 milliliters, which i'll just write as 0 .055 liters.

02:21 Oh, well, actually we could just convert this to grams because the next question is going to ask about grams...

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