25ml of a solution of na2c03 having a sepcific gravity of 1.25g solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N h2s04 that will be compltetly neutralized by 125g of na2c03 solution.
In first part I understand
109.5g/36.5g/mol = 3M
And then 3M x 0.0329L = 0.0987/2 0.04935
In second I do not still understand
How do get 470ml
Step by step explain
M Na2CO3 solution (the 25 mL) is
mols/L = 0.04935/0.025 = 1.974 M Na2CO3.
The 1.974M Na2CO3 now reacts with 0.84N H2SO4(0.84N = 0.42M). How much Na2CO3. That's 125 g of the solution. The density of that solution is 1.25 g/mL. volume = mass/density = 125g/1.25 g/mL = 100 mL Na2CO3.
Na2CO3 + H2SO4 ==> Na2SO4 + H2O + CO2
mols Na2CO3 = 0.100L x 1.974M = 0.1974.
mols H2SO4 = the same
M H2SO4 = mols H2SO4/L H2SO4. You have mols (0.1974) and you have M(0.84N or 0.42M). Solve for L H2SO4 That's 0.1974 mols/0.42 M = 0.470 L = 470 mL.
a 5 g mixture of natural gas containing ch4 and c2h4 was burnt in excess of oxygen yielding 14.5 g of co2 and h2o.What is the weight of ch4 and co2 in the mixture.
To calculate the volume of 0.84N H2SO4 that will be completely neutralized by 125g of Na2CO3 solution, we need to follow a series of steps. Let's go through each step:
Step 1: Determine the molar mass of Na2CO3.
The molar mass of Na2CO3 can be calculated by adding the atomic masses of its constituent elements:
Na: 22.99 g/mol (two atoms, so 22.99 x 2 = 45.98 g/mol)
C: 12.01 g/mol
O: 16.00 g/mol (three atoms, so 16.00 x 3 = 48.00 g/mol)
Adding the masses, the molar mass of Na2CO3 is: 45.98 + 12.01 + 48.00 = 105.99 g/mol.
Step 2: Calculate the number of moles of Na2CO3 in 125g.
Using the formula: Number of moles = mass / molar mass,
the number of moles of Na2CO3 = 125g / 105.99 g/mol = 1.179 mol.
Step 3: Determine the number of moles of H2SO4 required for neutralization.
The balanced chemical equation for the reaction between Na2CO3 and H2SO4 is:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
From the equation, we can see that two moles of H2SO4 are required to react with one mole of Na2CO3. Therefore, the number of moles of H2SO4 required is also 1.179 moles.
Step 4: Convert the normality (N) of H2SO4 to molarity (M).
Normality (N) is defined as the number of equivalents per liter of solution. Since H2SO4 is diprotic (can donate two protons), we need to convert it to molarity (M) using the formula:
Molarity = Normality / Number of equivalents.
For H2SO4, the number of equivalents is 2 (since it is diprotic).
0.84N H2SO4 = Molarity / 2
Molarity = 0.84N x 2 = 1.68M
Step 5: Calculate the volume of 1.68M H2SO4 solution required for neutralization.
Using the formula: Volume = moles / molarity,
Volume of H2SO4 = 1.179 moles / 1.68M = 0.701 L
Step 6: Convert the volume to milliliters.
1 L is equal to 1000 mL. So, 0.701 L = 0.701 x 1000 mL = 701 mL.
Therefore, the volume of 0.84N H2SO4 that will be completely neutralized by 125g of Na2CO3 solution is 701 mL.