Question

(Temperature is negative) How many liters of NO2 (at STP) can be produced from 149 grams of Cu reacting with concentrated nitric acid, HNO3 (aq)?

(The equation is balanced)


Cu + 4 HNO3 ⇒ Cu(NO3) 2 + 2 H2O + 2 NO2

63.55, 63.02, 187.57, 18.02, 46.01
^ g/mol

Answer 1

`\large \boxed{\text{106 L}}`

Step-by-step explanation:

We will need a balanced chemical equation with molar masses and volumes, so, let's gather all the information in one place.

MV/L: 22.71

M_r: 63.55

Cu + 4HNO₃ ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

m/g: 149

(a) Moles of Cu

`\text{Moles of Cu } =\text{149 g Cu } * \frac{\text{1 mol Cu }}{\text{63.55 g Cu }} =\text{2.344 mol Cu}`

(b) Moles of NO₂

The molar ratio is 2 mol NO₂:1 mol Cu

`\text{Moles of NO$_(2)$}= \text{2.344 mol Cu} * \frac{\text{2 mol NO$_(2)$}}{ \text{1 mol Cu}} = \text{4.689 mol NO$_(2)$}`

(c) Volume of NO₂

The volume of 1 mol of an ideal gas at STP (0 °C and 1 bar) is 22.71 L.

`\text{V} = \text{4.689 mol} * \frac{\text{22.71 L}}{\text{1 mol}} = \textbf{106 L}\\\\\text{You can produce $\large \boxed{\textbf{106 L}} $ of NO$_(2)$.}`