A 0.250 g chunk of sodium metal is cautiously dropped into amixture of 50.0 g of

Question

A 0.250 g chunk of sodium metal is cautiously dropped into amixture of 50.0 g of water and 50.0 g of ice, both @ 0 degreesCelsius. the reaction is 2Na(s) + 2H2O(l) --------> 2NaOH(aq) + H2(g) H = -368 kJ Will the ice met? Assume the final mixture has a specific heat capacity of 4.18J g -1 degrees C -1 Calculate the final temperature. A 0.250 g chunk of sodium metal is cautiously dropped into amixture of 50.0 g of water and 50.0 g of ice, both @ 0 degreesCelsius. the reaction is 2Na(s) + 2H2O(l) --------> 2NaOH(aq) + H2(g) H = -368 kJ Will the ice met? Assume the final mixture has a specific heat capacity of 4.18J g -1 degrees C -1 Calculate the final temperature.

Explanation / Answer

Heat of fusion for ice =333.55 J/g

Since there are 50 g ofice,
Heat required to melt the ice = 50 g * (333.55 J / 1 g)
                                           = 16677.5 J
.
Moles of sodium = mass/ atomic mass
                          = 0.25 g / 22.98 g/mol
                          = 0.0108 mol
.
2Na(s) + 2H2O (l) --------> 2NaOH(aq) + H2(g)
?H = -368 kJ
.
This is the heat given out for 2 moles of sodium
So heat given out by the reaction is = 0.0108 mol * (368 kJ / 2moles)
                                                     = 1.9872 kJ or 1987.2 J

Since this is less thanthe heat necessary to melt the ice, the ice does not melt.

Let final temperature beTf.
Heat = mass * specific heat * temperature difference
1987.2 J = 100 g * 4.18J/g.C * ( Tf - 0)
So final temperature, Tf = 4.75 C

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