Respuesta :
Explanation:
A) [tex]HCl(aq)+NaOH(aq)\rightarrow H_2O(l)+NaCl(aq)[/tex]
Mass of sodium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{40 g/mol}=0.0675 mol[/tex]
According to reaction , 1 mol of NaOH neutralizes with 1 mol of HCl.
Then 0.0675 mol of base will neutralize:
[tex]\frac{1}{1}\times 0.0675 mol=0.0675 mol[/tex] 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) [tex]2HNO_3(aq)+Ca(OH)_2(aq)\rightarrow 2H_2O(l)+Ca(NO_3)_2(aq)[/tex]
Mass of calcium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{57 g/mol}=0.04736 mol[/tex]
According to reaction , 2 mol of [tex]HNO_3[/tex] neutralizes with 1 mol of [tex]Ca(OH)_2[/tex].
Then 0.04736 mol of base will neutralize:
[tex]\frac{2}{1}\times 0.04736 mol=0.09472 mol[/tex] of [tex]HNO_3[/tex]
Mass of 0.09472 mol of [tex]HNO_3[/tex] :
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) [tex]H_2SO_4(aq)+2KOH(aq)\rightarrow 2H_2O(l)+K_2SO_4(aq)[/tex]
Mass of potassium hydroxide= 2.7 g
Moles of base = [tex]\frac{2.7 g}{56 g/mol}=0.04821 mol[/tex]
According to reaction , 1 mol of [tex]H_2SO_4[/tex] neutralizes with 2 mol of [tex]KOH[/tex].
Then 0.04821 mol of base will neutralize:
[tex]\frac{1}{2}\times 0.04821 mol=0.02410 mol[/tex] of [tex]H_2SO_4[/tex]
Mass of 0.02410 mol of [tex]H_2SO_4[/tex] :
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.
The amount, in grams, of each acid that would be needed for each of the reactions represented by the equations respectively, would be 2.46 grams, 4.59 grams, and 2.36 grams.
Stoichiometric calculations
From the first equation: HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)
The mole ratio of HCl to NaOH is 1:1.
Mole of 2.7 g NaOH = 2.7/40 = 0.0675 moles
Equivalent mole of HCl = 0.0675 moles
Mass of 0.0675 mole HCl = 0.0675 x 36.458 = 2.46 grams
For the second equation: 2HNO3(aq)+Ca(OH)2(aq)→2H2O(l)+Ca(NO3)2(aq)
Mole ratio of base to acid = 1:2
Mole of 2.7 grams Ca(OH)2 = 2.7/74.093 = 0.0364 moles
Equivalent mole of HNO3 = 0.0364 x 2 = 0.0728 moles
Mass of 0.0728 mole HNO3 = 0.0728 x 63.01 = 4.59 grams
For the third equation: H2SO4(aq)+2KOH(aq)→2H2O(l)+K2SO4(aq)
Mole ratio of acid to base = 1;2
Mole of 2.7 grams KOH = 2.7/56.1 = 0.0481 moles
Equivalent mole of H2SO4 = 0.0481/2 = 0.024 moles
Mass of 0.024 mole H2SO4 = 0.024 x 98.079 = 2.36 grams
More on stoichiometric calculations can be found here: https://brainly.com/question/8062886
