A 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 previous photograph, must be used to neutralise the acid? [Using NaOH: MM = 40 g/mol]
FIRST, one NEVER washes away spilled HCl with NaOH. Pellets of NaOH will cause more skin burns than 10% HCl every will. To neutralize acid spills use sodium hydrogen carbonate (baking soda) NaHCO3. Na2CO3 can be used also.
But to the problem.
NaOH + HCl --> NaCl + H2O
10% HCl (is that 10% HCl by volume or 10% HCl by mass). I will assume 10% by volume which means 10g HCl/100 mL solution. Therefore, in 530 mL we have 10g x (530/100) = estimated 53 g HCl which is
mol HCl = grams/molar mass = estimated 1.45
So you will need 1.45 mols NaOH.
g NaOH = mols NaOH x molar mass NaOH = ?