determine the quantity of NaOH whose concentration is 15% w/v to neutralize 176g H2SO4 , this is the exact problem and no its not an experiment. sorry to bother someone of you
Thanks. I can help with this. And the experiment part of my question meant that if an experiment I would need to know what you were doing.
As you have written it the problem now makes sense.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
First, how many moles NaOH do you have in the solution. That is 15% w/v which means 15 g NaOH/100 mL solution. moles = grams/molar mass = 15/40 = about 0.38 moles NaOH. You can redo this and be more exact. So the 15% w/v NaOH is about 0.38 moles/100 mL.
176 g H2SO4 is how many moles.
176/98 = about 1.8 moles. Convert moles H2SO4 to moles NaOH using the coefficients in the balanced equation. That will be 1.8 moles H2SO4 x (2 moles NaOH/1 mole H2SO4) = 1.8 x (2/1) = about 3.6 moles NaOH. Our NaOH has about 0.38 moles/100 mL, so how many mL will it take to obtain 3.6 moles. That will be 100 mL x (3.6 moles/0.38 moles) = about 947 mL of the 15% NaOH solution will be required. .
thanks a lot
To determine the quantity of NaOH needed to neutralize 176g of H2SO4 with a concentration of 15% w/v, we need to follow a series of steps. Here's how you can approach the problem:
Step 1: Convert the concentration of NaOH from w/v to molarity.
Since the concentration is given in terms of weight per volume (w/v), we need to convert it to molarity (M). To do this, we can use the formula:
Molarity (M) = (weight of solute in grams) / (molecular weight of solute × volume of solution in liters)
The molecular weight of NaOH is 40.00 g/mol. We will consider 100 mL of the 15% NaOH solution as the volume used.
Step 2: Calculate the molarity of the NaOH solution.
Let's assume that the density of the NaOH solution is 1.00 g/mL. Now we can calculate:
Volume of NaOH solution in liters = 100 mL × 1 L/1000 mL = 0.100 L
Molarity of NaOH = (0.15 g / 40.00 g/mol) / 0.100 L = 0.0375 mol/L
Step 3: Calculate the number of moles of H2SO4.
To determine the number of moles of H2SO4, we need to divide the given mass by the molar mass of H2SO4. The molar mass of H2SO4 is 98.09 g/mol.
Number of moles of H2SO4 = 176 g / 98.09 g/mol
Step 4: Find the stoichiometric ratio between NaOH and H2SO4.
Based on the balanced chemical equation for the neutralization reaction between NaOH and H2SO4, the stoichiometric ratio is 2:1. This means that 2 moles of NaOH react with 1 mole of H2SO4.
Step 5: Calculate the quantity of NaOH needed to neutralize the given H2SO4.
Using the stoichiometric ratio, we can determine the number of moles of NaOH needed to react with 176g of H2SO4:
Number of moles of NaOH = (Number of moles of H2SO4) × (2 moles of NaOH / 1 mole of H2SO4)
Once you calculate the moles of NaOH required, you can convert them to grams by multiplying them by the molar mass of NaOH.