If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
To determine the mass of chloride produced when heating potassium trioxochlorate V (KClO3) in the presence of MnO2, we need to consider the stoichiometry of the reaction and use the molar masses of the substances involved.
The balanced equation for the reaction can be written as:
2KClO3 + MnO2 → K2MnO4 + Cl2 + O2
From the equation, we can see that every 2 moles of KClO3 produce 1 mole of Cl2 (chlorine gas). We can use this ratio to calculate the moles of Cl2 produced.
First, determine the number of moles of KClO3:
Given mass of KClO3 = 15 grams
Molar mass of KClO3 = 39.1 + 35.5 + 3 * 16 = 122.5 g/mol
Moles of KClO3 = Mass / Molar mass
Moles of KClO3 = 15 g / 122.5 g/mol
Next, use the mole ratio between KClO3 and Cl2:
From the balanced equation, we know that 2 moles of KClO3 produce 1 mole of Cl2.
Moles of Cl2 = Moles of KClO3 / 2
Now, we can find the mass of Cl2 produced:
Mass of Cl2 = Moles of Cl2 * Molar mass of Cl2
The molar mass of Cl2 is 35.5 * 2 = 71 g/mol.
Finally, substitute the values to calculate the mass of Cl2 produced:
Mass of Cl2 = (15 g / 122.5 g/mol) / 2 * 71 g/mol
By performing the calculation, you will find the mass of chloride (Cl2) produced when heating 15 grams of potassium trioxochlorate V in the presence of MnO2.