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solong [7]
3 years ago
10

A 0.250 gram chunk of sodium metal is cautiously dropped into a mixture of 50.0 grams of water and 50.0 grams of ice, both at 0

degress celsius. The reaction is:2Na(s)+2H2O(l) yields 2NaOH(aq)+H2(g) Change in H is -368 KJWill the ice melt?
Assuming the final mixture has a specific heat capacity of 4.18 J/g*degrees celsius, calculate the final temperature. The enthalpy of fusion for ice is 6.02 KJ/mol.
Chemistry
1 answer:
JulsSmile [24]3 years ago
5 0

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.

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Pani-rosa [81]
<span>294400 cal The heating of the water will have 3 phases 1. Melting of the ice, the temperature will remain constant at 0 degrees C 2. Heating of water to boiling, the temperature will rise 3. Boiling of water, temperature will remain constant at 100 degrees C So, let's see how many cal are needed for each phase. We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion. 80 cal/g * 320 g = 25600 cal Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C 420 * 100 = 42000 cal Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away. 420 g * 540 cal/g = 226800 cal So the total number of cal used is 25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
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