A 0.250-g chunk of sodium metal is cautiously dropped into a mixture of 50.0 g of water and 50.0 g of ice, both at 08C. The reaction is 2Na (s) 1 2H2O (l) 88n 2NaOH (aq) 1 H2 (g) DH 5 2368 kJ

Question

Chemistry

Livia Moon

6 June, 21:22

A 0.250-g chunk of sodium metal is cautiously dropped into a mixture of 50.0 g of water and 50.0 g of ice, both at 08C. The reaction is 2Na (s) 1 2H2O (l) 88n 2NaOH (aq) 1 H2 (g) DH 5 2368 kJ

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Answers (1)

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Chico · Answer 1

The ice will not melt

Explanation:

Given:

Mass of chunk = 0.25 g

moles of sodium = mass / molar mass = 0.25 / 23 = 0.010 g

ΔH = - 368 KJ

Hence, 2 moles of Na produced 368 KJ

So 1 mole will produce 368 / 2 KJ and 0.01 mole will produce

= (368 / 2) * 0.01 = 2.0056 KJ

Also, moles of ice = mass / molar mass of water = 50 / 18 = 2.8 g

Energy required to melt the ice = enthalpy of fusion of ice * number of moles of ice

= 6.01 * 2.8 = 16.722 KJ

But we have 2.0056 KJ, so the ice will not melt.