4NaOH + 4HCI - NaCl + H20

NaOH + HCl —> NaCl + H₂O

Explanation:

From the question given above, we obtained the following equation:

4NaOH + 4HCl —> NaCl + H₂O

The equation can be balance as follow:

4NaOH + 4HCl —> NaCl + H₂O

There are 4 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NaCl as shown below:

4NaOH + 4HCl —> 4NaCl + H₂O

There are a total of 8 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of H₂O as shown below:

4NaOH + 4HCl —> 4NaCl + 4H₂O

Next, divide through by 4 to express the equation with the lowest coefficients.

NaOH + HCl —> NaCl + H₂O

Now the equation is balanced.

NOTE: Smallest coefficients are always used to represent a balanced equation.

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