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2 July, 20:17

42.6 g Cu are combined with 84.0 g HNO3 according to reaction:

3Cu + 8HNO3 _> 3Cu (NO3) 2 + 2NO + 4H2O

What reagent is limiting and how many grams of Cu (NO3) 2 are produced?

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  1. 2 July, 20:32
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    Mass of Cu (NO₃) ₂ produced = 125.67 g

    limiting reactant = Cu

    Explanation:

    Given dа ta:

    Mass of Cu = 42.6 g

    Mass of HNO₃ = 84.0 g

    What is limiting reactant = ?

    Mass of Cu (NO₃) ₂ produced = ?

    Solution:

    Chemical equation:

    3Cu + 8HNO₃ → 3Cu (NO₃) ₂ + 4H₂O + 2NO

    Number of moles of Cu:

    Number of moles = mass / molar mass

    Number of moles = 42.6 g/63.546 g/mol

    Number of moles = 0.67 mol

    Number of moles of HNO₃:

    Number of moles = mass / molar mass

    Number of moles = 84 g/63.01 g/mol

    Number of moles = 1.3 mol

    Now we will compare the moles of Cu (NO₃) ₂ with HNO₃ and Cu from balance chemical equation.

    HNO₃ : Cu (NO₃) ₂

    8 : 3

    1.3 : 3/8*1.3 = 6

    Cu : Cu (NO₃) ₂

    3 : 3

    0.67 : 0.67

    Number of moles of Cu (NO₃) ₂ produced by Cu are less so Cu will limiting reactant.

    Mass of Cu (NO₃) ₂:

    Mass = number of moles * molar mass

    Mass = 0.67 mol * 187.56 g/mol

    Mass = 125.67 g
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