Plz help w/calorimatry

Question

The following acid-base reaction is performed in a coffee cup calorimeter:

H (aq) OH-(aq) --> H2O(l)

The temperature of 110 g of water rises from 25.0C to 26.2C when 0.10 mol of H is reacted with 0.10 mol of OH-.

Calculate qwater

Calculate ΔH for the reaction

Calculate ΔH if 1.00 mol OH- reacts w 1 mol H

plz plz show ALL steps

Answer 1

Hi,

q = (sp. heat)mΔt

q= 4.18 (J / g°C) (110 g)(26.6°C - 25.0°C)

q= 550 J  <~~ Answer #1

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ΔH = -(q) = - 550 J <~~Answer #2

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We know that when 0.010 mol of H+ or OH- reacts, ΔH is - 550 J:

Therefore, for 1.00 mol:

ΔH = (1.00 mol H+) (-550 J / 0.010 mol H+)

ΔH = -5.5 x 104 J    <~~Answer #3

Hope this helps.