Question

helppppppppppppppppppppWhat quantity of heat is associated with the synthesis of 45.0 g of NH3, according to the following AH = +1170 kj 4 NO(g) + 6 H₂O(1)→ 4 NH3(g) + 5 O₂(g) equation

Answer 1

774.265 kJ

Step-by-step explanation

Given reaction:

4 NO(g) + 6 H₂O(1) → 4 NH3(g) + 5 O₂(g) ΔH = +1170 kJ

The given reaction is an endothermic reaction (ΔH = +1170 kJ) in which 4 moles of production of ammonia (NH3) requires 1170 kJ amount of heat.

From the Periodic Table, each mole of ammonia corresponds to 17grams.

This implies that (4 mol x 17 g/mol) = 68 grams of NH3 requires 1170 kJ amount of energy.

To find the quantity of heat energy associated with the production of 45.0 grams of NH3, we need to calculate the heat energy associated per gram of NH3 by as follows:

`\frac{1170\text{ kJ}}{68\text{ grams}}=17.20588235\text{ }kJ\text{/}g`

Therefore, the quantity of heat associated with the synthesis of 45.0 g of NH3 is:

`45.0g*17.20588235\text{ }kJ\text{/}g=774.265\text{ }kJ`