For the reaction shown calculate how many grams of each...

For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to form products. Assume that there is more than enough of the other reactant. 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) 11.0g Na2CO3 A) m NaCl = B) m H2O = C) m CO2 =

Transcript

00:01 Okay, we have this reaction here of hcl with sodium carbonate and we make sodium chloride water and carbon dioxide and we have 11 grams of na2cl and we're told that this is excess so we don't want to care about what what what what how much acl we have we have 11 grams of this and that's all we need to worry about we want to know how many grams of each of these we're going to make from this 11 grams of an 82 c03 is assuming a 100 % reaction or complete reaction.

00:36 And so let's convert, we'll need the molar mass for each other species.

00:41 If we're going into moles and then out of moles for all four of these species.

00:44 So let's go ahead find the molar mass for all these.

00:46 So sodium has a molar mass at 22 .99.

00:52 Chloride has a molar mass at 35 .45.

00:56 Corbin has one of 12 .01.

00:59 Oxygen has one of 16.

01:01 And hydrogen has one of 1 .01.

01:04 So i pulled all these from the parc table.

01:08 And so now we can start piecing them together to form these compounds.

01:12 So in a 2 co3, that's going to be 22 .99 times 2 plus 12 .01 plus 16 times 3.

01:28 And so this comes out to 105 .99 grams per mole.

01:35 So that's what na2co3 is.

01:39 In acl it's going to be 22 .99 plus 35 .45.

01:48 This comes out to 58 .44.

01:53 Water is h2o, which comes out to 16 plus 2 times 1 .01 or 18 .02.

02:02 And lastly, co2 comes out to 16 times 2 for the oxygens plus 12 .101 equals 44 .0.

02:13 1.

02:14 All these are grams per mole.

02:21 So now let's convert our 11 grams of sodium carbonate into moles that we can then use to find how many moles of each of these species we make...

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