530 mL solution 10% HCl has been spilt onto the bench. It cannot be diluted or washed away. How much of the granular NaOH from the container seen in a must be used to neutralise the acid? [Using NaOH: MM = 40 g/mol]
1. I have NaOH + HCl --> NaCl + H2O.
10% HCI in a 530ml solution equals to 53grams/molar mass. Therefore 53 g/mol divided by 40g/mol Naoh = 1.325g/mol of NaOH to neutralise the HCL as 1.325g/mol x 40mm of NaOH equals 53g and therefore will neutralise the HCL (53g/mm) as HCI and NaOH have a 1.1 mole ratio.
Is this correct if not how do I convert it to grams?
thankyou guys.
No it isn't. Go back and look at that work I did for you last night. You have smatterings right but you've incorrectly interpreted what I wrote in several places.
mols HCl = 53/36.5 = ?
mols NaOH = same as mols HCl because of the 1:1 ratio.
g NaOH = mols NaOH x molar mass NaOH.
The 1.325 is not right either. You have 53 g HCl and not 53 g/mol.
To calculate how much NaOH is needed to neutralize the HCl, you can follow the steps below:
1. Determine the amount of HCl in the 530 mL solution. Since the solution is 10% HCl, you can multiply the volume (530 mL) by the concentration (10%) to find the amount in grams:
530 mL * 10% = 53 grams of HCl
2. Convert the grams of HCl to moles. To do this, divide by the molar mass of HCl:
53 grams / 36.5 g/mol = 1.45 moles of HCl
3. Calculate the moles of NaOH needed to neutralize the HCl. Since the balanced equation between NaOH and HCl is 1:1, the moles of NaOH needed will also be 1.45 moles.
4. Finally, convert the moles of NaOH to grams by multiplying by the molar mass of NaOH:
1.45 moles * 40 g/mol = 58 grams of NaOH
Therefore, approximately 58 grams of NaOH from container A must be used to neutralize the spilled HCl solution on the bench.