How much heat in kilocalories is absorbed in the reaction of 1.94g of Na with H2O? 2Na(s)+2H2O(l)->2NaOH(aq)+H2(g) change in heat= -88.0kcal/mol
If dH is - heat is released and not absorbed. Is that kcal/mol or kcal/2 mols.
That is how my question is worded. How much heat (in kilocalories) is evolved or absorbed in the reaction of 1.94g of Na with H2O? change in heat= -88.0 kcal/mol
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To find out how much heat is absorbed in the reaction between 1.94g of Na and H2O, we need to use the molar mass of Na and conversions between mass and moles.
1. First, we need to calculate the number of moles of Na. To do this, divide the mass (in grams) of Na by its molar mass, which is 22.99 g/mol.
Moles of Na = 1.94g Na / 22.99 g/mol = 0.084 mol Na
2. Next, we can use the balanced equation to determine the molar ratio between Na and the change in heat:
2 Na(s) + 2 H2O(l) -> 2 NaOH(aq) + H2(g)
The change in heat is given as -88.0 kcal/mol of reaction.
-88.0 kcal/mol represents the heat change for 2 moles of Na reacting. Therefore, to find the heat change for 0.084 mol of Na, we need to calculate the proportion:
Heat change = (-88.0 kcal/mol) * (0.084 mol Na / 2 mol Na)
3. Now we can calculate the heat change:
Heat change = -88.0 kcal/mol * 0.042 = -3.696 kcal
Therefore, approximately -3.696 kilocalories (kcal) of heat is absorbed in the reaction of 1.94g of Na with H2O.