If 30 g of HCl is reacted with excess NaOH, and 10g of NaCl is produced, what is the percent yield of the experiment?

Percent yield of NaCl = 20.85%

Explanation:

Given data:

Mass of HCl = 30 g

Actual yield of NaCl = 10 g

Percent yield of NaCl = ?

Solution:

Chemical equation:

HCl  + NaOH   →    NaCl  + H₂O

Moles of HCl:

Number of moles = mass/ molar mass

Number of moles = 30 g/ 36.5 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of HCl with NaCl

                             HCl        :        NaCl

                                1          :            1

                              0.82      :        0.82

Theoretical yield:

Mass = number of moles × molar mass

Mass = 0.82 mol × 58.5 g/mol

Mass = 47.97 g

Percentage yield:

Formula:

Percent yield = Actual yield / theoretical yield × 100

Now we will put the values.

Percent yield = 10 g / 47.97 g × 100

Percent yield = 20.85%

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