How many grams of CI2 will be produced when 2.59 mol MnO2 react with excess HCI?
To find the number of grams of Cl2 produced, we need to use stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between MnO2 and HCl is:
4 HCl + MnO2 -> MnCl2 + 2 H2O + Cl2
According to the balanced equation, 1 mole of MnO2 reacts with 4 moles of HCl to produce 1 mole of Cl2.
Given that we have 2.59 mol of MnO2, we can multiply it by the mole ratio to find the number of moles of Cl2 produced:
2.59 mol MnO2 x (1 mol Cl2 / 1 mol MnO2) = 2.59 mol Cl2
To convert the moles of Cl2 to grams, we need to use the molar mass of Cl2.
The molar mass of Cl2 is calculated by adding the atomic masses of chlorine (Cl):
Molar mass of Cl2 = (2 x atomic mass of Cl) = (2 x 35.45 g/mol) = 70.9 g/mol
Now, we can calculate the number of grams of Cl2 produced:
2.59 mol Cl2 x (70.9 g Cl2 / 1 mol Cl2) = 183.531 g Cl2
Therefore, when 2.59 mol of MnO2 reacts with excess HCl, 183.531 grams of Cl2 will be produced.