Calculate the maximum mass of potassium sulfate, K2SO4, that could be formed when 4.9 g of sulfuric acid reacts with excess potassium hydroxide. H2SO4 + 2KOH → K2SO4 + 2H2O

From 4.9 g of sulfuric acid 8.175 g of potassium sulfate is formed.

Explanation:

Given data:

Mass of H₂SO₄ = 4.9 g

Mass of K₂SO₄ formed = ?

Solution:

Chemical reaction:

H₂SO₄ + 2KOH    →    K₂SO₄ + 2H₂O

Number of moles of H₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 4.9 g/ 98.1 g/mol

Number of moles = 0.05 mol

Now we will compare the moles of  H₂SO₄ with  K₂SO₄ .

                       H₂SO₄         :             K₂SO₄

                            1             :                  1

                       0.05           :             0.05

Mass of K₂SO₄:

Mass = number of moles × molar mass

Mass = 0.05 mol × 174.3 g/mol

Mass = 8.715 g

Thus, from 4.9 g of sulfuric acid 8.175 g of potassium sulfate is formed.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol