Chemistry, 28.08.2019 19:30 rosemarybooker

How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(ii) nitrate solution reacts with excess aluminum metal? show all of the work needed to solve this problem. . 2 al (s) + 3 fe(no3)2 (aq) 3 fe (s) + 2 al(no3)2 (aq)

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The balanced chemical reaction is:

2 Al (s) + 3 Fe(NO3)2 (aq) = 3 Fe (s) + 2 Al(NO3)2 (aq)

The ratios or the coefficients between the reactants and the products will be used for further calculations.

305 g Fe(NO3)2 solution (.755) = 230.275 g Fe(NO3)2

230.275 g Fe(NO3)2 ( 1mol / 179.87 g) ( 3 mol Fe / 3 mol Fe(NO3)2) ( 55.85 g / mol) = 71.500 g Fe

Answer from: queeneaf

We are given with 305 grams of a 75.5 percent by mass iron(II) nitrate solution to produce metal iron upon reaction with aluminum. The mass of iron(II) nitrate is 305 grams * 0.755 or equal to 230.275 grams. In this case, 1 mole of iron(II) nitrate is equal to 1 mole of iron. Thus we just divide the amount of iron(II) nitrate with molar mass of iron(II) nitrate and multiply with the molar mass of iron. The answer is 71.51 grams.

Answer from: Quest

answer; natural resources;

Answer from: Quest

definition

entropy is a measure of the randomness or the extent of disorder of a chemical process.

enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

entropy is measured in joule per kelvin (jk⁻¹) whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

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