Question

What volume (in L) of 1M HCl is needed to react completely with 0.6 grams of calcium metal?

Suppose that a temperature rise of 11 degrees Celcius was observed when 80.0 mL of 1.00 M HCl was mixed with 20.0 mL of 4.00 sodium hydroxide. What is the molar heat of neutralization?

Answer 1

Answer on Question#38976 - Chemistry - Inorganic Chemistry

Task:

What volume (in L) of 1M HCl is needed to react completely with 0.6 grams of calcium metal?

Solution:

The chemical equation of the reaction is

$$\mathrm{Ca} + 2\mathrm{HCl} = \mathrm{CaCl}_2 + \mathrm{H}_2$$

The number of moles of Ca is

$$n(\mathrm{Ca}) = \frac{m(\mathrm{Ca})}{MW(\mathrm{Ca})} = \frac{0.6}{40} = 0.015 \text{ mol}$$

the number of moles of HCl is twice the number of moles of Ca

$$n(\mathrm{HCl}) = \frac{2n(\mathrm{Ca})}{2} = \frac{2 \cdot 0.015}{0.030 \text{ mol}}$$

Now we can find the volume of HCl solution

$$C(\mathrm{M}) = \frac{n(\text{mol})}{V(\mathrm{L})}, \text{ where } C \text{ is molar concentration}$$ $$V(\mathrm{L}) = \frac{n(\text{mol})}{C(\mathrm{M})} = \frac{0.030}{1} = 0.030 \mathrm{L}$$

Answer: $V(\mathrm{L}) = 0.030 \mathrm{L}$

Task:

Suppose that a temperature rise of 11 degrees Celsius was observed when $80.0\,\mathrm{mL}$ of $1.00\,\mathrm{M}$ HCl was mixed with $20.0\,\mathrm{mL}$ of 4.00 sodium hydroxide. What is the molar heat of neutralization?

Solution:

The chemical equation of the reaction is

$$\mathrm{HCl} + \mathrm{NaOH} = \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$$

The number of moles of acid is

$$C(\mathrm{M}) = \frac{n(\text{mol})}{V(\mathrm{L})}$$ $$n(\mathrm{HCl}) = \frac{C(\mathrm{HCl}) \cdot V(\mathrm{L})}{1.00 \cdot 0.080} = 0.080 \text{ mol}$$

The number of moles of base is

$$C(\mathrm{M}) = \frac{n(\text{mol})}{V(\mathrm{L})}$$ $$n(\mathrm{NaOH}) = \frac{C(\mathrm{NaOH}) \cdot V(\mathrm{L})}{4.00 \cdot 0.020} = 0.080 \text{ mol}$$

The mass of acid is

$$m(\mathrm{HCl}) = \frac{n(\text{mol}) \cdot M(\text{mol})}{0.080 \cdot 36.5} = 2.92 \mathrm{g}$$

The molar heat of neutralization is

$$Q = \frac{m \cdot C_p \cdot \Delta T}{m}$$

- the mass of acid

$C_p$ - The specific heat capacity of the aqueous solutions is 4.184\,\mathrm{J\,^{\circ}C^{-1}\,g^{-1}}

$$Q = \frac{m \cdot C_p \cdot \Delta T}{2.92 \cdot 4.184 \cdot 11} = 134.4\,\mathrm{J}$$

To calculate the molar heat of neutralization we must divide the obtained value by the number of moles reacted:

$$Q_m = \frac{Q}{n} = \frac{134.4}{0.08} = 1680\,\mathrm{J/mol}$$

Answer: $Q_m = 1680\,\mathrm{J/mol}$