Asked by Alex
How much energy is released when 2.50g of NH3 rxs with excess N20 given the following 2NH3+3N2O -> 4N2_3H2O; DELTA(H)=-1010KJ.
Please give work and setup :)
Please give work and setup :)
Answers
Answered by
DrBob222
Look up the delta H<sup>o</sup><sub>f</sub> values for each, then
dHrxn = (n*dHf products) - (n*dHf reactants). This dHrxn will be for 34 g NH3. For 2.5 g it will be
dHrxn x (2/34) = ?
dHrxn = (n*dHf products) - (n*dHf reactants). This dHrxn will be for 34 g NH3. For 2.5 g it will be
dHrxn x (2/34) = ?
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