How many milliliters of 18 M H2SO4 are required to react with 250 mL of 2.5 M Al(OH)3 if the products are aluminum sulfate and water?
Here is an example.
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52.0 mL H2SO4
To determine the volume of 18 M H2SO4 required to react with 250 mL of 2.5 M Al(OH)3, we first need to find the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between H2SO4 and Al(OH)3 can be written as:
3 H2SO4 + 2 Al(OH)3 -> Al2(SO4)3 + 6 H2O
From the equation, we can see that 3 moles of H2SO4 react with 2 moles of Al(OH)3 to produce 1 mole of Al2(SO4)3 and 6 moles of water.
Next, we need to determine the number of moles of Al(OH)3 in 250 mL of 2.5 M solution. This can be calculated using the formula:
Moles = Concentration (M) x Volume (L)
Moles of Al(OH)3 = 2.5 M x 0.250 L
= 0.625 moles
Since the stoichiometric ratio between H2SO4 and Al(OH)3 is 3:2, we can calculate the number of moles of H2SO4 required using the equation:
Moles of H2SO4 = (2/3) x Moles of Al(OH)3
= (2/3) x 0.625 moles
= 0.417 moles
Finally, we need to calculate the volume of 18 M H2SO4 required to obtain 0.417 moles. This can be done using the formula:
Volume (L) = Moles / Concentration (M)
Volume of 18 M H2SO4 = 0.417 moles / 18 M
= 0.023 L
= 23 mL
Therefore, 23 mL of 18 M H2SO4 is required to react with 250 mL of 2.5 M Al(OH)3.
To determine the amount of 18 M H2SO4 required to react with 250 mL of 2.5 M Al(OH)3, we can use a balanced chemical equation to establish the molar ratio between the reactants.
The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and aluminum hydroxide (Al(OH)3) is:
2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O
From the balanced equation, we can see that the molar ratio between Al(OH)3 and H2SO4 is 2:3. This means that for every 2 moles of Al(OH)3, 3 moles of H2SO4 are required.
First, we need to determine the number of moles of Al(OH)3 present in 250 mL of 2.5 M solution.
Moles of Al(OH)3 = volume (L) x molarity (mol/L)
Moles of Al(OH)3 = 0.250 L x 2.5 mol/L
Moles of Al(OH)3 = 0.625 mol
Next, we can use the molar ratio from the balanced equation to determine the number of moles of H2SO4 required.
Moles of H2SO4 = (moles of Al(OH)3 x Stoichiometric coefficient H2SO4) / Stoichiometric coefficient Al(OH)3
Moles of H2SO4 = (0.625 mol x 3) / 2
Moles of H2SO4 = 0.9375 mol
Finally, we need to convert the moles of H2SO4 to milliliters by utilizing the molarity of the H2SO4 solution.
Volume (mL) of H2SO4 = moles of H2SO4 x volume (mL) / moles
Volume (mL) of H2SO4 = 0.9375 mol x 1000 mL / 18 mol
Volume (mL) of H2SO4 = 52.083 mL
Therefore, approximately 52.083 mL of 18 M H2SO4 are required to react with 250 mL of 2.5 M Al(OH)3.