What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed as shown in the following reaction? (NO CALCULATOR) Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) (A) 13.5 g (B) 27.0 g (C) 40.0 g (D) 67.0 g

13.5

Explanation:

Molar mass of Fe2O3 = 2*56 + 16*3 = 112 + 48 =160 approximate value

Number of moles of Fe2O3 = mass/molar mass = 40/160 = 0.25 mol

The reaction is balanced. Fine

1 mol of Fe2O3> 2 mol of Al

0.25 mol> 0.25*2 = 0.5 mol of Al

Mass of Al required = Number of moles * Molar mass

= 0.5 * 27 = 13.5 g

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.