For each of the following acid-base reactions, calculate how many grams of each acid are necessary to completely react with and neutralize 2.7 g of the base. part a : hcl(aq)+naoh(aq)→h2o(l)+nacl(aq)par t b : 2hno3(aq)+ca(oh)2(aq)→2h2o(l)+ca(no 3)2(aq)part c : h2so4(aq)+2koh(aq)→2h2o(l)+k2so4(aq )

Explanation:

A)  

Mass of sodium hydroxide= 2.7 g

Moles of base =

According to reaction , 1 mol of NaOH neutralizes with 1 mol of HCl.

Then 0.0675 mol of base will neutralize:

of HCl.

Mass of 0.0675 mol of HCl =  0.0675 mol × 35.5 g/mol = 2.396 g

2.396 grams of acid will completely react with and neutralize 2.7 g of the sodium hydroxide.

B)  

Mass of calcium hydroxide= 2.7 g

Moles of base =

According to reaction , 2 mol of neutralizes with 1 mol of .

Then 0.04736 mol of base will neutralize:

of

Mass of 0.09472 mol of :

0.09472 mol × 63g/mol = 5.9673 g

5.9673 grams of acid will completely react with and neutralize 2.7 g of the calcium hydroxide.

C)  

Mass of potassium hydroxide= 2.7 g

Moles of base =

According to reaction , 1 mol of neutralizes with 2 mol of .

Then 0.04821 mol of base will neutralize:

of

Mass of 0.02410 mol of :

0.02410 mol × 98 g/mol = 2.3618 g

2.3618 grams of acid will completely react with and neutralize 2.7 g of the potassium hydroxide.