Question
15.0g of potassium trioxochlorate(v) was crushed and heated with about 0.1g of manganese(iv)oxide Calculate the mass of KCl that will be produced assuming the reaction was complete
Answers
Your name for KClO3 is not correct and is not an accepted name by IUPAC. Manganese(IV) oxide is a correct name for MnO2.That iv should be capitalized.
2KClO3 ==> 2KCl + 3O2
mols KClO3 = grams/molar mass = 15.0/molarmass KClO3 = ?
Convert to mols KCl produced. That is mols KClO3 x (2 mols KCl/2 mols KClO3) = ?
Then grams KCl = mols KCl x molar mass KCl = ?
Post your work if you get stuck.
2KClO3 ==> 2KCl + 3O2
mols KClO3 = grams/molar mass = 15.0/molarmass KClO3 = ?
Convert to mols KCl produced. That is mols KClO3 x (2 mols KCl/2 mols KClO3) = ?
Then grams KCl = mols KCl x molar mass KCl = ?
Post your work if you get stuck.
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