Answer:
369.20 grams of NaOH are required to produce 9.23 moles of NaCl.
Step-by-step explanation:
To answer this question, we need to know the balanced chemical equation for the reaction between NaOH and HCl, assuming that the reaction takes place in an aqueous solution:
NaOH + HCl → NaCl + H2O
According to this equation, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.
To calculate the amount of NaOH required to produce 9.23 moles of NaCl, we need to use the mole ratio between NaOH and NaCl from the balanced equation. Since the mole ratio is 1:1, we know that we need 9.23 moles of NaOH to produce 9.23 moles of NaCl.
The molar mass of NaOH is 40.00 g/mol, which means that 1 mole of NaOH weighs 40.00 grams. Therefore, 9.23 moles of NaOH weighs:
9.23 moles x 40.00 g/mol = 369.20 grams of NaOH
So, 369.20 grams of NaOH are required to produce 9.23 moles of NaCl.