Calculate the volume (in mL) of 9.0M H2SO4 that you would need to convert 0.070 moles of KAl(OH)4 to K2SO4 and Al2(SO4)3 according to the following equation:
2 KAl(OH)4 + 4 H2SO4->K2SO4 + Al2(SO4)3 + 8 H2O
Follow the same procedure as for the alum problems. You may need an additional formula that is M = moles/L for the H2SO4.
.07moleKAL(oh4)*2mole(h2so4)/9mole(h2so4)=15.6mlH2So4
To calculate the volume of 9.0M H2SO4 needed to convert 0.070 moles of KAl(OH)4, we need to use the stoichiometry of the balanced equation.
From the balanced equation:
2 moles of KAl(OH)4 react with 4 moles of H2SO4
We can set up a proportion using the moles of KAl(OH)4:
2 moles KAl(OH)4 / 4 moles H2SO4 = 0.070 moles KAl(OH)4 / x moles H2SO4
Cross-multiplying and solving for x:
2 * x = 4 * 0.070
2x = 0.280
x = 0.140 moles H2SO4
Next, we need to calculate the volume of 9.0M H2SO4 needed to contain 0.140 moles of H2SO4.
Molarity (M) is defined as moles per liter (mol/L). Therefore, we can use the equation:
Moles of solute = Molarity * Volume (in liters)
Rearranging the equation to solve for Volume:
Volume (in liters) = Moles of solute / Molarity
Plugging in the values:
Volume (in liters) = 0.140 moles H2SO4 / 9.0M
Since the question asks for the answer in milliliters (mL), we need to convert liters to milliliters:
1 liter = 1000 milliliters
Volume (in mL) = Volume (in liters) * 1000
Plugging in the values and calculating:
Volume (in mL) = (0.140 moles H2SO4 / 9.0M) * 1000
Volume (in mL) = 15.6 mL
Therefore, you would need approximately 15.6 mL of 9.0M H2SO4 to convert 0.070 moles of KAl(OH)4 to K2SO4 and Al2(SO4)3.
To calculate the volume of 9.0M H2SO4 needed to convert 0.070 moles of KAl(OH)4, we need to use the balanced chemical equation and the concept of stoichiometry.
In the balanced chemical equation:
2 KAl(OH)4 + 4 H2SO4 -> K2SO4 + Al2(SO4)3 + 8 H2O
We can see that it takes 4 moles of H2SO4 to react with 2 moles of KAl(OH)4. Therefore, the stoichiometric ratio of H2SO4 to KAl(OH)4 is 4:2, which simplifies to 2:1.
To convert moles to volume, we need to use the molarity (M) and the formula:
Moles = Volume (in liters) x Molarity
To find the volume in liters, we rearrange the formula:
Volume (in liters) = Moles / Molarity
Let's calculate the volume:
1. Convert the moles of KAl(OH)4 to moles of H2SO4 using the stoichiometry:
0.070 moles KAl(OH)4 x (4 moles H2SO4 / 2 moles KAl(OH)4) = 0.140 moles H2SO4
2. Calculate the volume of 9.0M H2SO4 needed by dividing the moles by the molarity:
Volume (in liters) = 0.140 moles / 9.0 M = 0.0156 liters
3. Convert the volume from liters to milliliters:
Volume (in milliliters) = 0.0156 x 1000 = 15.6 mL
Therefore, you would need approximately 15.6 mL of 9.0M H2SO4 to convert 0.070 moles of KAl(OH)4 to K2SO4 and Al2(SO4)3 according to the given equation.