Question

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO3)2, how many grams of Cd(OH)2 will be formed in the following precipitation reaction?

2 NaOH(ag) + Cd(NO)2(ag) -> Cd(OH)2 (s)
+ 2 NaNOs(aq)

Answer 1

84.31 grams of Cd(OH)2 will be formed.

Step-by-step explanation:

To find the grams of Cd(OH)2 formed, we need to use stoichiometry. First, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product formed.

Let's start by calculating the moles of NaOH and Cd(NO3)2:

Moles of NaOH = mass / molar mass = 20.0 g / 40.0 g/mol = 0.5 mol

Moles of Cd(NO3)2 = concentration x volume = 1.00 M x 0.750 L = 0.75 mol

According to the balanced equation, the ratio of NaOH to Cd(NO3)2 is 2:1. Therefore, we can see that Cd(NO3)2 is the limiting reactant because it has fewer moles.

Now, let's use the stoichiometry to calculate the moles of Cd(OH)2 formed:

From the balanced equation, we know that 1 mole of Cd(NO3)2 reacts to form 1 mole of Cd(OH)2. Therefore, the moles of Cd(OH)2 formed is also 0.75 mol.

Finally, we can calculate the grams of Cd(OH)2 formed:

Grams of Cd(OH)2 = moles x molar mass = 0.75 mol x 112.41 g/mol = 84.31 g