Scan_Doc0014 - 3.11 The Concept of Limiting Reagent 107 2 ~ What are the moles of NaHC03 in 1.00 g To find the moles of NaHC03 we need to know the

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Unformatted text preview: 3.11 The Concept of Limiting Reagent 107 2 ~ What are the moles of NaHC03 in 1.00 g? To find the moles of NaHC03, we need to know the molar mass (84.01 g/mol). 1.00 g...NaHe()3 X 1 mol NaHC03 84.01 g..NaHee)3 = 1.19 X 10-2 mol NaHC03 in the balanced equation? 3 ~ What is the mole ratio between HCl and NaHC03 1 mol HCI 1 mol NaHC03 4 ~ What are the moles of HCl? 1.19 X 10-2 ~ X 1~ 1 mol HCI = 1.19 X 10-2 mol HCI Thus 1.00 g of NaHC03 will neutralize 1.19.X 10-2 mol HCI. For Mg(OH)2 1 ~ What is the balanced equation? Mg(OHMs) + 2HCl(aq) ~ 2H20(l) + MgCI2(aq) 2 ~ What are the moles of Mg(OHh in 1.00 g? To find the moles of Mg(OHh, we need to know the molar mass (58.32 g/mol). 1.00 ~ X 1 mol Mg(OH)2 ~ 58.32 2 . = 1.7l X 10-2 mol Mg(OH)2 3 ~ What is the mole ratio between HCI and Mg(OH)2 in the balanced equation? 2 mol HCI 1 mol Mg(OH)2 4 ~ What are the moles of HCI? 1.71 X 10-2 ~ X 1 Milk of magnesia contains of Mg(OHh(s). a suspension 2 mol HCI ~ 2 = 3.42 X 10-2 mol HCI Thus 1.00 g of Mg(OH)2 will neutralize 3.42 X 10-2 mol HCI. • Since 1.00 g NaHC03 neutralizes 1.19 X 10-2 mol HCI and 1.00 g Mg(OH)2 neutralizes 3.42 X 10-2 mol HCI, Mg(OH)2 is the more effective antacid. SEE EXERCISES 3.101 AND 3.102 3.11 ~ The Concept of Limiting Reagent Suppose you have a part-time job in a sandwich shop. One very popular sandwich is always made as follows:' 2 slices bread + 3 slices meat + 1 slice cheese ~ sandwich Assume that you come to work one day and find the following quantities of ingredients: 8 slices bread 9 slices meat 5 slices cheese How many sandwiches can you make? What will be left over? ...
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This note was uploaded on 12/13/2010 for the course CHEM 2301 taught by Professor Bill during the Spring '10 term at South Texas College.

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