2.) Striking a match stick on the matchbox involves the reaction of potassium chlorate and red phosphorus with the following equation: 10 KCIO3 (s) + 3 P4 (s) ---> 3 P4O10 (s) + 10 KCI (s) (A) How much reactant (KCIO3) will be needed to produce 105.8 g of P4O10? (B) How much reactant (P4) will be needed to produce 105. g of P4010? Molar Masses: 1 mole of KCIO3 = 122.55 grams of KCIO3 1 mole of P4 = 123.88 grams of P4 1 mole of P4O10 = 283.88 grams of P4010 Ratio: 10 mole of KCIO3 = 3 mole of P4010 3 mole of P4 = 3 mole of P4010

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2.) Striking a match stick on the matchbox involves the
reaction of potassium chlorate and red phosphorus with
the following equation:
10 KCIO3 (s) +3 Р4(s) ---> 3 РАО10 (s) + 1О КСІ (5)
(A) How much reactant (KCIO3) will be needed to produce
105.8 g of P4010?
(B) How much reactant (P4) will be needed to produce 105.8
g of P4010?
Molar Masses:
1 mole of KCIO3 = 122.55 grams of KCIO3
%3D
1 mole of P4 = 123.88 grams of P4
%3D
1 mole of P4010 = 283.88 grams of P4010
Ratio:
10 mole of KCIO3 = 3 mole of P4010
3 mole of P4 = 3 mole of P4O10
Transcribed Image Text:2.) Striking a match stick on the matchbox involves the reaction of potassium chlorate and red phosphorus with the following equation: 10 KCIO3 (s) +3 Р4(s) ---> 3 РАО10 (s) + 1О КСІ (5) (A) How much reactant (KCIO3) will be needed to produce 105.8 g of P4010? (B) How much reactant (P4) will be needed to produce 105.8 g of P4010? Molar Masses: 1 mole of KCIO3 = 122.55 grams of KCIO3 %3D 1 mole of P4 = 123.88 grams of P4 %3D 1 mole of P4010 = 283.88 grams of P4010 Ratio: 10 mole of KCIO3 = 3 mole of P4010 3 mole of P4 = 3 mole of P4O10
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