Question

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

Answer 1

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

Answer 2

Answer:

Strontium hydroxide.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Sr(OH)_2+H_2CO_3\rightarrow SrCO_3+2H_2O$

Thus, the first step is to compute the moles of both strontium hydroxide (molar mass = 121.63 g/mol) and carbonic acid given the mass and volume and concentration respectively:

$n_{Sr(OH)_2}=12.5gSr(OH)_2*\frac{1molSr(OH)_2}{121.63gSr(OH)_2} =0.103molSr(OH)_2\\\\n_{H_2CO_3}=3.5mol/L*0.150L=0.525molH_2CO_3$

Therefore, as they are in a 1:1 molar ratio in the chemical reaction, the same moles of both strontium hydroxide and carbonic acid must react, nevertheless, there are more moles of carbonic acid, for that reason, the limiting reactant is strontium hydroxide.

Best regards.