#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# HNO3 used as a reagent in stoichiometric analysis has specific gravity of 1.5 and contains 63% HNO3 by mass. The volume of magnesium hydroxide solution which is 2.9 g/L in Mg(OH)2 required for neutralizing 20 mL of above HNO3 solution is?

14 Points
9 years ago

2HNO3+Mg(OH)2=Mg(NO3)2+2H2O.

Mass of HNO3 taken is 1.5*20/1000 gm = 0.03 gm.

Actual mass of HNO3  present 63% of 0.03 gm = 0.0189 gm = 0.0189/63 moles = 0.0003 mole.

Moles of H+ present = 0.0003 .

Moles of OH- needed (from Mg(OH)2) = 0.0003.

So, Moles of Mg(OH)2 needed = 0.00015 = 0.008745 gm Mg(OH)2 .

Let volume needed be v mL.

Then, 2.9*v/1000 = 0.008745

=> v = 3.01.

Ans:3.01 g of 100% pure Mg(OH)2 solution will be needed.