Chemistry question

Question

What volume (mL) of 1.50 × 10^-1 M hydrochloric acid can be neutralized with 105 mL of 0.500 M sodium hydroxide? Include the balanced equation with your work. 

Answer 1

HCl(aq) + NaOH(aq) ==> NaCl(aq) + H2O(l) .. balanced equation

How many moles of NaOH are present?

105 ml x 1 L / 1000 ml x 0.500 mol / L = 0.0525 moles NaOH present

How many moles of HCl are needed to neutralize this?

From the balanced equation above, 1 mole HCl reacts with 1 mole NaOH, so 0.0525 moles HCl needed

What volume of 1.50x10^-1 M HCl will contain the required 0.0525 moles of HCl?

0.0525 moles x 1 L / 1.50x10^-1 moles = 0.350 L = 350. mls HCl needed (3 sig. figs.)