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#15 How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate 2KClO 3  2KCl + 3O 2 6.54 g KClO 3 (given) ? molecules.

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Presentation on theme: "#15 How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate 2KClO 3  2KCl + 3O 2 6.54 g KClO 3 (given) ? molecules."— Presentation transcript:

1 #15 How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate 2KClO 3  2KCl + 3O g KClO 3 (given) ? molecules produced 6.54 g KClO 3 X 1 1 mole KClO 3 X g KClO 3 3 moles O 2 2 mole KClO 3 x 6.02 x molecules O 2 1 mole O 2 = 4.82 x molecules O = molar mass of KClO 3

2 #16 The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. 3NO 2 + H 2 O  2HNO 3 + NO ? g NO x molecules NO (given) 5.00 x molecules NO x 1 1 mole NO x 6.02 x molecules NO 3 moles NO 2 x 1 mole NO 46 g NO 2 1 mole NO 2 = 11.5 g NO = 46 molar mass of NO 2

3 #17 The equation for the combustion of carbon monoxide is: 2CO + O 2  2CO 2 How many liters of Oxygen are required to burn 3.86 L of carbon monoxide? 2CO + O 2  2CO L CO? L O L CO x 1 1 mole CO x 22.4 L CO 1 mole O 2 x 2 mole CO 22.4 L O 2 1 mole O 2 = 1.93 L O 2

4 #18 Phosphorus and hydrogen can be combined to form phosphine (PH 3 ) P 4(s) + 6H 2(g)  4PH 3(g) How many liters of phosphine are formed when 0.42 L of hydrogen reacts with phosphorus? P 4(s) + 6H 2(g)  4PH 3(g) ? L PH L H L H 2 x 1 1 mole H 2 x 22.4 L H 2 4 moles PH 3 x 6 moles H L PH 3 1 mole PH 2 = 0.28 L PH 3

5 #19 Consider this equation: CS 2 + 3O 2  CO 2 + 2SO 2 ? L SO mL O 2 (the only trick here is to convert to liters from mL and the rest of the problem is the same) 27.9 mL O 2 x 1 1 L O 2 x 1000 mL O 2 1 mole O 2 x 22.4 L O 2 2 moles SO 2 x 3 moles O L SO 2 1mole SO 2 =.018 L SO mL SO 2

6 #20 Consider this equation: CS 2 + 3O 2  CO 2 + 2SO 2 ? dL CO L SO 2 (the only trick here is to convert to deciliters from Liters) 0.38 L SO 2 x 1 1mole SO 2 x 22.4 L SO 2 1 mole CO 2 x 2 moles SO L CO 2 x 1 moles CO 2 10 dL CO 2 1 L CO 2 = 1.9 dL CO 2.19 L CO 2 1 dL is 100 ml or 1/10 of a liter


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