What volume of .109M HNO3, in milliliters is required to react completely with 2.5g of Ba(OH)2?

2HNO3 (aq) + Ba(OH)2 -----> 2H2O (l) + Ba(NO3)2 (Aq)

Walk me through this please.

first convert 2.5g Ba(OH)2 into mol

the mm is 137.3+ (17x2) = 171.3 g / mol

so 2.5g of Ba(OH2) is 1.46*10^-2 mol


From the equation you know that for every 2 moles of HNO3 you produce 1 mole of Ba(OH)2, 2:1 ratio

If you have 1.46*10^-2 mol of Ba(OH)2 that means you'll need double the amount of HNO3 or 2.92*10^-2 mol of HNO3

original question gives you a molarity of 0.109M

Molarity = moles solute / volume (liters) of solution

or M = m / V

so

V = m / M
= 2.92*10^-2 / 0.109
= 2.67 liters

Thank you very much

This post was edited by westonn on Dec 1 2013 10:14pm