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Striking a match stick on the matchbox involves the reaction of potassium chlorate and red phosphorus with the following equation: 10 KCIO3 (s) + 3P4 (s) → 3P4O10 (s) + 10 KCI (s) How much of each reactant will be needed to produce 105.8 g of tetraphosphorus decaoxide (P4O10)?​

Respuesta :

Answer:

m(P4) =  46.175 (grams)

m (KClO3) = 149 (grams)


Explanation:

1) n(P4) = n(P4O10);

m(P4)/M(P4) = m(P4O10)/M(P4O10);

m(P4) = M(P4)*m(P4O10)/M(P4O10)
          = 123.90*105.8/283.89
          = 46.175 (grams)

2) Analogously, 10n(P4O10) = 3n(KClO3)

m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
                = 10*122.55*105.8/283.89/3
                = 149 (grams).