A piece of aluminium weighing 2.7 gm is heated with 75 ml of H2SO4(specific gravity 1.18 containing 24.7 percent H2SO4
by weight.)After the metal is carefully dissolved,the solution is dissolved to 400ml.Calculate tne molarity of the free H2SO4 in the resulting solution.(IIT 1992)
2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2
mols Al = g/atomic mass
mols H2SO4 = M x L. L = 0.075 and M = 1.18 x 1000 x 0.247 x (1/98) = ?
From the problem I suppose we assume H2SO4 is the excess reagent. Use stoichiometry to determine H2SO4 used, subtract from the initial amount to find that remaining, the M of the remaining -= mols/L solution.
Post your work if you get stuck.
A 25-mL sample of 0.160M solution of NaOH is titrated with 17 mL of an unknown solution of H2SO4. What is the molarity of the sulfuric acid solution? A. 0.004M H2SO4 B. 0.235M H2SO4 C. 0.117M H2SO4 D. 0.002M H2SO4
To calculate the molarity of the free H2SO4 in the resulting solution, we need to first determine the number of moles of H2SO4 present.
Step 1: Calculate the moles of H2SO4 in the initial 75 ml of solution.
Given that the specific gravity of the H2SO4 solution is 1.18 and it contains 24.7% H2SO4 by weight, we can find the density of the solution:
Density = specific gravity * density of water
Density = 1.18 * 1 g/cm³ (since the density of water is 1 g/cm³)
Density = 1.18 g/cm³
Next, we calculate the mass of H2SO4 in the 75 ml solution:
Mass of H2SO4 = volume of solution * density of solution
Mass of H2SO4 = 75 ml * 1.18 g/cm³
Mass of H2SO4 = 88.5 g (rounded to 1 decimal place)
Now, we calculate the moles of H2SO4 in the 75 ml solution:
Moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4
The molar mass of H2SO4 = (2 * molar mass of H) + molar mass of S + (4 * molar mass of O)
The molar mass of H = 1 g/mol, molar mass of S = 32.06 g/mol, and molar mass of O = 16 g/mol.
Molar mass of H2SO4 = (2 * 1) + 32.06 + (4 * 16)
Molar mass of H2SO4 = 98.09 g/mol (rounded to 2 decimal places)
Moles of H2SO4 = 88.5 g / 98.09 g/mol
Moles of H2SO4 = 0.9029 mol (rounded to 4 decimal places)
Step 2: Calculate the moles of H2SO4 in the final 400 ml solution.
Since the solution is dissolved to 400 ml, the moles of H2SO4 is not changed during this step.
Step 3: Calculate the molarity of the free H2SO4 in the resulting solution.
Molarity (M) = Moles of solute (H2SO4) / Volume of solution (in liters)
Volume of solution = 400 ml = 0.4 L (since 1 L = 1000 ml)
Molarity = 0.9029 mol / 0.4 L
Molarity = 2.26 M (rounded to 2 decimal places)
Therefore, the molarity of the free H2SO4 in the resulting solution is 2.26 M.
To calculate the molarity of the free H2SO4 in the resulting solution, we need to follow a step-by-step approach:
Step 1: Calculate the moles of aluminum used.
To do this, divide the mass of aluminum (2.7 g) by the molar mass of aluminum (26.98 g/mol).
Moles of aluminum = 2.7 g / 26.98 g/mol
Step 2: Calculate the moles of H2SO4 initially present in 75 ml of the sulfuric acid solution.
First, calculate the weight of H2SO4 present in 75 ml solution.
Weight of H2SO4 = Specific gravity x Volume x % of H2SO4 by weight
Weight of H2SO4 = 1.18 x 75 ml x (24.7/100)
Next, calculate the moles of H2SO4.
Moles of H2SO4 = Weight of H2SO4 / Molar mass of H2SO4
Molar mass of H2SO4 = 2(1.008 g/mol) + 32.06 g/mol + 4(16.00 g/mol)
Moles of H2SO4 = Weight of H2SO4 / Molar mass of H2SO4
Step 3: Calculate the final volume of the solution after dissolution.
The solution volume increases from 75 ml to 400 ml after the dissolution of the aluminum. Hence, the final volume is 400 ml.
Step 4: Calculate the molarity of the free H2SO4.
To calculate the molarity of the H2SO4, divide the moles of H2SO4 by the final volume in liters.
Molarity (M) = Moles of H2SO4 / Final volume in liters
Final volume in liters = 400 ml / 1000 ml/L
Now, plug in the values and calculate the molarity of the free H2SO4 in the resulting solution.