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STOICHIMETRY Limiting Reagents AMY MARKS. What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl.

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Presentation on theme: "STOICHIMETRY Limiting Reagents AMY MARKS. What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl."— Presentation transcript:

1 STOICHIMETRY Limiting Reagents AMY MARKS

2 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? Mg + HCL -> MgCl2 + H2

3 First step? Balance Equation End it all now Call Mr. Kirk What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? Mg + HCL -> MgCl2 + H2

4 CORRECT! The balanced equation is now Mg + 2HCl -> MgCl2 + H2 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation?

5 You’re getting there…What to do now? Convert grams to moles Put given over 1 Find the STP Mg + 2HCl -> MgCl2 + H2

6 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 50.0g Mg 1 x ? 24.3g Mg 1 mol Mg 24.3g Mg 6.02x10^23 1 mol Mg Mg + 2HCl -> MgCl2 + H2

7 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 50.0g Mg 1 x 1 mol Mg 24.3g Mg x ? 1 mol H2 1 mol Mg 1 mol H2 1 mol Mg 22.4 L Mg + 2HCl -> MgCl2 + H2

8 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 50.0g Mg 1 x 1 mol H2 1 mol Mg xx? 22.4 L 1 mol H2 Just stop  1 mol H L Mg + 2HCl -> MgCl2 + H2 1 mol Mg 24.3g mg

9 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 50.0g Mg x 1 mol Mg x 1 mol H2 x 22.4 L g Mg 1 mol Mg 1 mol H2 = ? 46.1 L H2 461 L H L H2 Mg + 2HCl -> MgCl2 + H2

10 You’re not done.. Continue on to 2 nd half of the problemContinue on to 2 nd half of the problem.

11 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 1 75g HCl 1 75g HCl 1 mol What is the first step? Mg + 2HCl -> MgCl2 + H2

12 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 75g HCl 1 x ? 1 mol HCl 36.5g HCl 1 mol HCl 1 mol Mg 1 mol HCl

13 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 75g HCl x 1 mol HCl x ? g HCl 2 mol HCl __________ 1 mol H2 __________ 2 mol HCl 1 mol H2 __________ 1 Mg + 2HCl -> MgCl2 + H2

14 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 75g HCl x 1 mol HCl x 1 mol H2 x ? g HCl 2 mol HCl 22.4 L 1 mol H2 1 mol H L Click me Mg + 2HCl -> MgCl2 + H2

15 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? 75g HCl x 1 mol HCl x 1 mol H2 x 22.4 L H2 = ? g HCl 2 mol HCl 1 mol H L H223 L H2 Mg + 2HCl -> MgCl2 + H2

16 YEAHHHH BUDDY you’re almost done. Close enough. Let’s go. What is the excess reagent of the equation?

17 Which do you pick? 23 L 50.0g 75g fail

18 EXCESS REAGENT cool, you got it. 23 L H2 x ? 1 1 mol H L H L H2 1 mol H2 Who cares. Mg + 2HCl -> MgCl2 + H2

19 Excess Reagent Cont.. 23L H2 x 1 mol H2 x ? L H2 Mg + 2HCl -> MgCl2 + H2 1 mol ____ 2 mol 1 mol H2 ____ 1 mol Mg 1 mol Mg ____ 1 mol H2

20 23L H2 x 1 mol H2 x 1 mol Mg x ? L H2 1 mol H2 WOOO. Got it right… Mg + 2HCl -> MgCl2 + H g 1 mol Mg 2 mol Mg 44.6 g 1 mol Mg 24.3 g

21 Almost finished! Excess Reagent cont.. 23L H2 x 1 mol H2 x 1 mol Mg x 24.3Mg = ? L H2 1 mol H2 1 mol Mg Mg + 2HCl -> MgCl2 + H g 24 g 25 g

22 In order to find the excess reagent you must now subtract the previous total from the amount that was not the limited reagent. Previous total : 25g Limited reagent : HCl Non limited : Mg What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl according to the following equation? Mg + 2HCl -> MgCl2 + H = =50

23 FINALLY DONE The excess reagent is 25.0

24 How Depressing..

25 So close 

26 YEAHHH..no.

27 Sorry, but..no.

28 U mad? Back to start.  Stay mad.


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