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5 December, 01:07

Instructions:Select the correct answer. If 155 grams of potassium (K) reacts with 122 grams of potassium nitrate (KNO3), what is the limiting reagent? 2KNO3 + 10K 6K2O + N2

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Answers (2)
  1. 5 December, 05:00
    0
    2KNO₃ + 10K = 6K₂O + N₂

    n (K) = m (K) / M (K)

    n (K) = 155g/39.1g/mol=3.964 mol

    n (KNO₃) = m (KNO₃) / M (KNO₃)

    n (KNO₃) = 122g/101.1g/mol=1.207 mol

    K:KNO₃ = 10:2 = 5:1

    3.964:1.207=3.3:1

    K is the limiting reagent
  2. 5 December, 05:00
    0
    31310 / 390 = > 80.28 g of KNO3 (K is limiting reagent)
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