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LR Practice

# LR Practice - EXAMPLE Zinc and sulfur react to form zinc...

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EXAMPLE Zinc and sulfur react to form zinc sulfide, a substance used in phosphors that coat the inner surfaces of TV picture tubes. The equation for the reaction is Zn + S -----> ZnS In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react.

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EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) Which is the limiting reactant? (b) How many grams of ZnS can be formed from this particular reaction mixture? (c) How many grams of which element will remain unreacted in this experiment?
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn ( 12.0 g Zn ) (1 mol Zn) ( 1 mol ZnS ) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------- (65.38 g Zn) ( 1 mol Zn ) (1 mol ZnS)

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EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn (12.0 g Zn) (1 mol Zn) (1 mol ZnS) (97.5 g ZnS) #g ZnS = -------------------------------------------------------------- (65.38 g Zn) (1 mol Zn) (1 mol ZnS) = 17.9 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S ( 6.50 g S ) (1 mol S) ( 1 mol ZnS ) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------- (32.06 gS ) ( 1 mol S ) (1 mol ZnS)

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EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S (6.50 g S) (1 mol S) (1 mol ZnS) (97.5 g ZnS) #g ZnS = ----------------------------------------------------------- (32.06 gS ) (1 mol S) (1 mol ZnS) = 19.8 g ZnS
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.9 g ZnS can be produced .

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EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced. (c) How many grams of which element will remain unreacted in this experiment? if use all 12.0 g Zn
EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced. (c) if use all 12.0 g Zn ( 12.0 g Zn ) (1 mol Zn) ( 1 mol S ) (32.06 g S) #g S = -------------------------------------------------------- (65.38 g Zn) ( 1 mol Zn ) (1 mol S)

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EXAMPLE In a particular experiment 12.0 g of Zn are mixed with 6.50 g S and allowed to react. Zn + S -----> ZnS (a) The limiting reactant? (b) #grams of ZnS? if use all 12.0 g Zn #g ZnS = 17.9 g ZnS if use all 6.50 g S #g ZnS = 19.8 g ZnS Therefore, Zn is the limiting reactant and 17.8 g ZnS can be produced.
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LR Practice - EXAMPLE Zinc and sulfur react to form zinc...

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