QuestionJuly 22, 2025

9. How much heat will be transferred when 14.9 g of ammonia reacts with excess oxygen according to the following equation? Circle one: Is this reaction endothermic or exothermic? Additionally, rewrite the thermochemical equation in the other format. 4NH_(3)+5O_(2)arrow 4NO+6H_(2)O Delta H=-1170kJ

9. How much heat will be transferred when 14.9 g of ammonia reacts with excess oxygen according to the following equation? Circle one: Is this reaction endothermic or exothermic? Additionally, rewrite the thermochemical equation in the other format. 4NH_(3)+5O_(2)arrow 4NO+6H_(2)O Delta H=-1170kJ
9. How much heat will be transferred when 14.9 g of ammonia reacts with excess oxygen according to the following equation? Circle one: Is this reaction endothermic or exothermic? Additionally, rewrite the thermochemical equation in the other format. 4NH_(3)+5O_(2)arrow 4NO+6H_(2)O Delta H=-1170kJ

Solution
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Answer

-255.94 kJ (exothermic) Explanation 1. Calculate moles of ammonia Molar mass of NH_3 is 17.03 g/mol. Moles = \frac{14.9 \text{ g}}{17.03 \text{ g/mol}} = 0.875 \text{ mol}. 2. Determine heat transfer per mole From the equation, 4 moles of NH_3 release 1170 kJ. Heat per mole = \frac{-1170 \text{ kJ}}{4} = -292.5 \text{ kJ/mol}. 3. Calculate total heat transferred Total heat = 0.875 \text{ mol} \times (-292.5 \text{ kJ/mol}) = -255.9375 \text{ kJ}. 4. Identify reaction type Since \Delta H is negative, the reaction is exothermic. 5. Rewrite thermochemical equation 4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g) \quad \Delta H = -1170 \text{ kJ}

Explanation

1. Calculate moles of ammonia<br /> Molar mass of $NH_3$ is 17.03 g/mol. Moles = $\frac{14.9 \text{ g}}{17.03 \text{ g/mol}} = 0.875 \text{ mol}$.<br /><br />2. Determine heat transfer per mole<br /> From the equation, 4 moles of $NH_3$ release 1170 kJ. Heat per mole = $\frac{-1170 \text{ kJ}}{4} = -292.5 \text{ kJ/mol}$.<br /><br />3. Calculate total heat transferred<br /> Total heat = $0.875 \text{ mol} \times (-292.5 \text{ kJ/mol}) = -255.9375 \text{ kJ}$.<br /><br />4. Identify reaction type<br /> Since $\Delta H$ is negative, the reaction is exothermic.<br /><br />5. Rewrite thermochemical equation<br /> $4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g) \quad \Delta H = -1170 \text{ kJ}$
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