If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
Potassium trioxochlorate (V) is another name for KClO3. The balanced reaction is
2KClO3(s) + heat �¨ 2KCl(s) + 3O2(g)
------------ (MnO2 catalyst)
15 g of KClO3 is 15g/122.6 g/mol = 0.1224 mol KClO3. You will make the same number of moles of KCl. Multiply that by the molar mass of KCl (74.6 g/mol) for the answer.
I get 9.13 moles
To determine the mass of chloride produced when heating potassium trioxochlorate V in the presence of MnO2, we need to understand the chemical reaction taking place and use stoichiometry.
The balanced chemical equation for the reaction is:
2KClO3 + MnO2 -> 2KCl + MnO4
From the equation, we can see that 2 moles of KClO3 produce 2 moles of KCl. This means there is a 1:1 ratio between moles of KClO3 and moles of KCl.
To find the mass of KCl, we need to determine the number of moles of KCl produced. We use the molar mass of KClO3 to convert the mass of KClO3 to moles.
The molar mass of KClO3 is:
K (39.10 g/mol) + Cl (35.45 g/mol) + 3O (16.00 g/mol) = 122.55 g/mol
Using the molar mass, we can convert the mass of KClO3 to moles:
15 g KClO3 * (1 mol KClO3 / 122.55 g KClO3) = 0.1224 mol KClO3
Since the ratio between KClO3 and KCl in the balanced equation is 1:1, we can conclude that 0.1224 moles of KCl are produced.
Next, we convert moles of KCl to mass using the molar mass of KCl:
K (39.10 g/mol) + Cl (35.45 g/mol) = 74.55 g/mol
The mass of KCl produced is:
0.1224 mol KCl * (74.55 g KCl / 1 mol KCl) = 9.128 g KCl
Therefore, the mass of chloride produced when 15 grams of potassium trioxochlorate V is heated in the presence of MnO2 is approximately 9.128 grams.