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?

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