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Question #81918

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in an unknown volume of water (where X is a hypothetical metal):

X + 2 H2O → X(OH)2 + H2

In this process, the water temperature rose from 25.0 °C to 37.1 °C. If 0.00579 mol of "X" was consumed during the reaction, and the ΔH of this reaction with respect to the system is -1278 kJ mol-1 , what volume of water (in mL) was present in the calorimeter?

The specific heat of water is 4.184 J g-1 °C-1

Expert's answer

Answer on Question #81918, Chemistry / Organic Chemistry

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in an unknown volume of water (where X is a hypothetical metal):

X + 2 H_2O \rightarrow X(OH)_2 + H_2

In this process, the water temperature rose from $25.0^{\circ}\mathrm{C}$ to $37.1^{\circ}\mathrm{C}$ . If 0.00579 mol of "X" was consumed during the reaction, and the $\Delta H$ of this reaction with respect to the system is -1278 kJ mol⁻¹, what volume of water (in mL) was present in the calorimeter?

The specific heat of water is $4.184\ \mathrm{J\ g^{-1}\ °C^{-1}}$

Solution

We should find the heat that was released by the reaction:

X + 2 H_2O \rightarrow X(OH)_2 + H_2 \quad \Delta H = -1278\ \mathrm{kJ/mol}

1278 kJ is released when 1 mole of "X" reacts with water

x kJ is released when 0.00579 mol of "X" react with water

Solve the proportion:

\frac{1278}{x} = \frac{1}{0.00579}

x = 7.39962

Q_{\text{released}} = 7.39962\ \mathrm{kJ}

Q_{\text{related}} = Q_{\text{absorbed by water}}

Q_{\text{absorbed by water}} = \mathrm{cm\ \Delta T}

7.39962 \times 10^3\ \mathrm{J} = 4.184\ \mathrm{J/g \cdot °C} \cdot m_{\text{water}} \cdot (37.1^{\circ}\ \mathrm{C} - 25.0^{\circ}\mathrm{C})

m_{\text{water}} = 146.16\ \mathrm{g}

d_{\text{water}} = 1\ \mathrm{g/cm^3}

V = m/d

V_{\text{water}} = 146.16\ \mathrm{g} / 1\ \mathrm{g/cm^3} = 146.16\ \mathrm{cm^3} = 146.16\ \mathrm{mL}

Question #81918

Expert's answer

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