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
![pH=-log[H^{+}]=-log[H_{3}O]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D%3D-log%5BH_%7B3%7DO%5D%20%20)
. Using the value in formula, we get:![pH=-log[1.8*10^{-4}]=3.745](https://tex.z-dn.net/?f=pH%3D-log%5B1.8%2A10%5E%7B-4%7D%5D%3D3.745%20)