The following thermochemical equation is for the reaction of hydrogen peroxide(l

ID: 840472 • Letter: T

Question

The following thermochemical equation is for the reaction of hydrogen peroxide(l) to form water(l) and oxygen(g).

How many grams of H2O2(l) would have to react to produce 29.5 kJ of energy?

_______grams

The following thermochemical equation is for the reaction of nitrogen monoxide(g) with hydrogen(g) to form nitrogen(g) and water(l).

When 10.2 grams of nitrogen monoxide(g) react with excess hydrogen(g)_______, kJ of energy are_________.evolved/absorbed.

The following thermochemical equation is for the reaction of nitrogen monoxide(g) with oxygen(g) to form nitrogen dioxide(g).

How many grams of NO(g) would have to react to produce 22.4 kJ of energy?

______grams

(1) 2 H2O2(l) => 2 H2O(l) + O2(g), DH = -196 kJ

Moles of H2O2 = energy produced/(-DH) x 2

= 29.5/196 x 2 = 0.3010 mol

Mass of H2O2 = moles x molar mass of H2O2

= 0.3010 x 34.015

= 10.2 g

(2) 2 NO(g) + 2 H2(g) => N2(g) + 2 H2O(l), DH = -752 kJ

Moles of NO = mass/molar mass of NO

= 10.2/30.006 = 0.3399 mol

Since DH is negative => energy is evolved

Energy evolved = moles of NO/2 x (-DH)

= 0.3399/2 x 752

= 128 kJ of energy are evolved

(3) 2 NO(g) + O2(g) => 2 NO2(g), DH = -114 kJ

Moles of NO = energy produced/(-DH) x 2

= 22.4/114 x 2 = 0.3930 mol

Mass of NO = moles x molar mass of NO

= 0.3930 x 30.006

= 11.8 g

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