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How many grams of nitrogen monoxide are present if 33.7 L of ammonia and 41.3 L of Oxygen are reacted at 296.3 K and 1.26 atm? What is the enthalpy change for the reaction? NH3(g) + O2(g) -> NO(g) + H2O(l)

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Answer:

29.7g of NO are present

Explanation:

With volume, absolute temperature and pressure we can determine moles of ammonia and oxygen that reacts. With these moles we can determine limiting reactant and moles (And mass) of nitrogen monoxide produced as follows:

Moles Ammonia:

PV = nRT

PV/RT = n

1.26atm*33.7L / 0.082atmL/molK*296.3K = 1.748 moles ammonia

Moles Oxygem:

PV = nRT

PV/RT = n

1.26atm*41.3L / 0.082atmL/molK*296.3K = 2.142 moles oxygen

Based on the chemical reaction, 1 mole of NH3 reacts per mole of O2 producing 1 mole of NO.

Thus, limiting reactant is ammonia and moles of NO produced are = Moles of ammonia added = 1.748mol NH3

Mass Ammonia:

1.748mol NH3 * (17g/mol) =

29.7g of NO are present