When heated, KClO 3 decomposes into KCl and O 2 . 2 KClO 3 ⟶ 2 KCl + 3 O 2 If this reaction produced 71.1 g KCl , how many grams of O 2 were produced?

The correct answer is 45.9 g O₂

Explanation:

The descomposition reaction is the following:

2 KClO₃ → 2 KCl + 3 O₂

From this,, we know that 2 mol KClO₃ produce 2 mol of KCl and 3 mol of O₂. Now, we can use the molar mass of the compounds and rewrite the equation with the mass of each compound:

MM KClO₃= 39 g/mol + 35.4 g/mol + (3 x 16 g/mol)= 122.4 g/mol

2 KClO₃ = 2 mol x 122.4 g/mol= 244.8 g

MM KCl= 39 g/mol + 35.4 g/mol= 74.4 g/mol

2 KCl= 2 mol x 74.4 g/mol= 148.8 g

MM O₂= 2 x 16 g/mol= 32 g/mol

3 O₂= 3 mol x 32 g/mol= 96 g

With this, we can write a recipe in mass:

2 KClO₃   →    2 KCl  +    3 O₂

244.8 g    →   148.8 g +  96 g

244.8 g of KClO₃ produce 148.8 g. If we have 71.1 g KCl, the mass of KClO₃ from which this mass of KCl is produced is:

244.8 g KClO₃ 148.8 g KCl

          x             71.1 g KCl

x= 71.1 g KCl x 244.8 g KClO₃/148.8 g KCl= 116.97 g KClO₃

Finally, we calculate the mass of O₂ produced by 116.97 g of KClO₃ taking into account that the recipe says that 244.8 g of KClO₃ produce 96 g of O₂.

244.8 g KClO₃ 96 g O₂

116.97 g KClO₃ x= 116.97 g KClO₃ x 96 g O₂/244.8 g KClO₃

x= 45.87 g O₂= 45.9 g O₂

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying

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