Question

What mass of Cu(NO₃)2 can be prepared?

Answer 1

The mass of Cu(NO₃)₂ that can be prepared depends on the quantity of reactants used, following stoichiometric principles and balanced chemical equations.

Explanation:

To determine the mass of Cu(NO₃)₂ that can be prepared, one needs to consider the balanced chemical equation and the molar masses of the reactants and products involved. For instance, if the reaction involves the reaction of copper oxide (CuO) with nitric acid (HNO₃), the balanced equation is:

`\[ \text{CuO} + 2\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + \text{H}_2\text{O} \]`

By determining the limiting reactant (the reactant that limits the product's formation) based on the given quantities of reactants and their stoichiometry, the theoretical yield of Cu(NO₃)₂ can be calculated. This calculation involves converting the given mass of the limiting reactant to the expected mass of the product using stoichiometry and molar ratios.

Using stoichiometry and the molar mass of Cu(NO₃)₂, one can calculate the theoretical yield. For example, if 50 grams of CuO reacts with an excess of HNO₃, given the molar mass of Cu(NO₃)₂ as X g/mol, the calculation would be:

`\[ \text{Moles of CuO} = \frac{\text{Given mass}}{\text{Molar mass of CuO}} \]`

`\[ \text{Moles of Cu(NO}_3)_2 = \text{Moles of CuO} * \frac{\text{Moles of Cu(NO}_3)_2}{\text{Moles of CuO}} \]`

`\[ \text{Mass of Cu(NO}_3)_2 = \text{Moles of Cu(NO}_3)_2 * \text{Molar mass of Cu(NO}_3)_2 \]`

This calculation will provide the maximum theoretical yield of Cu(NO₃)₂ that can be obtained from the given amount of CuO. Factors like reaction efficiency and practical limitations may result in the actual yield being lower than the theoretical yield.