According to the following reaction, how many moles of water will be formed upon the complete reaction of 22.6 grams of hydrochloric acid with excess iron(III) oxide?
To determine the number of moles of water formed in the reaction, we first need to balance the equation, if it hasn't been provided to us. The balanced equation for the reaction between hydrochloric acid (HCl) and iron(III) oxide (Fe2O3) is:
Fe2O3 + 6HCl → 2FeCl3 + 3H2O
Now, let's calculate the number of moles of hydrochloric acid (HCl) using its molar mass.
1. Find the molar mass of HCl:
The molar mass of hydrogen (H) is approximately 1 g/mol, and the molar mass of chlorine (Cl) is approximately 35.5 g/mol. Therefore, the molar mass of HCl is:
1 g/mol (H) + 35.5 g/mol (Cl) = 36.5 g/mol
2. Calculate the number of moles of HCl:
Given: Mass of HCl = 22.6 g
Number of moles of HCl = Mass of HCl ÷ Molar mass of HCl
Number of moles of HCl = 22.6 g ÷ 36.5 g/mol = 0.62 mol (rounded to two decimal places)
Since there is an excess of iron(III) oxide, it means that all the hydrochloric acid will react completely. According to the balanced equation, for every 6 moles of HCl, 3 moles of water are formed. Therefore, we can use stoichiometry to find the number of moles of water.
3. Calculate the number of moles of water:
Given: Number of moles of HCl = 0.62 mol
Number of moles of water = (0.62 mol HCl) × (3 mol H2O ÷ 6 mol HCl)
Number of moles of water = 0.31 mol (rounded to two decimal places)
Therefore, upon the complete reaction of 22.6 grams of hydrochloric acid with excess iron(III) oxide, 0.31 moles of water will be formed.