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(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2 NaNO₃(aq)

Respuesta :

Answer:

[tex]m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]

Explanation:

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In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

[tex]n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2[/tex]

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

[tex]m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2[/tex]

Best regards.

The mass of  Cd(OH)₂ that would be formed is 36.6 g

From the question,

We are to determine the mass of Cd(OH)₂ that would be formed

The given balanced chemical equation for the reaction is  

2NaOH(aq) + Cd(NO₃)₂(aq) → Cd(OH)₂ (s) + 2NaNO₃(aq)

This means

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂ and 2 moles of NaNO₃

Now, we will determine the number of moles of each reactant present

  • For NaOH

Mass = 20.0 g

Molar mass = 39.997 g/mol

From the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Number of moles of NaOH present = [tex]\frac{20.0}{39.997 }[/tex]

Number of moles of NaOH present = 0.50 mole

  • For Cd(NO₃)₂

Volume = 0.750 L

Concentration = 1.00 M

From the formula

Number of moles = Concentration × Volume

∴ Number of moles of Cd(NO₃)₂ present = 1.00 × 0.750

Number of moles of Cd(NO₃)₂ present = 0.750 mole

Since,

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂

Then,

0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂

∴ The number of moles of Cd(NO₃)₂ that reacted is 0.25 mole

Now,

From the balanced chemical reaction

2 moles of NaOH reacts with 1 mole of Cd(NO₃)₂ to produce 1 mole of Cd(OH)₂

Then,

0.5 mole of NaOH will react with 0.25 mole of Cd(NO₃)₂ to produce 0.25 mole of Cd(OH)₂

∴ Number of moles of Cd(OH)₂ formed is 0.25 mole

Now, for the mass of Cd(OH)₂ that would be formed

From the formula

Mass = Number of moles × Molar mass

Molar mass of Cd(OH)₂ = 146.43 g/mol

∴Mass of Cd(OH)₂ formed = 0.25 × 146.43

Mass of Cd(OH)₂ formed = 36.6075 g

Mass of Cd(OH)₂ formed≅ 36.6 g

Hence, the mass of  Cd(OH)₂ that would be formed is 36.6 g

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