Imagine you only have 100 mL of acetic acid. How could you determine how much baking soda to add to get the largest amount of gas possible?

Explanation:

You need the concentration of the acetic acid.

With the concentration you can find the number of moles of acid in the given 100mL of acetic acid solution.

By stoichiometry of the reaction, you determine the number of moles of baking soda:

CH₃COOH(aq) + NaHCO₃(aq)  → CH₃COONa(aq) + CO₂(g)  + H₂O(aq)

        ↑                        ↑                                                   ↑

acetic acid         baking soda                                        gas

From that balanced molecular equation, you find that 1 mole of acetic acid react with 1 mole of baking soda.

Then, using the molar mass of bakind soda, you convert the number of moles into mass in grams:

Then, adding that mass of baking soda, may be a little excess to be sure, all the acetic acid would be consumed getting the largest amount of gas possible.

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

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