1. Using the balanced equation, answer the following questions:

2KClO3 → 2KCl + 3O2

a. How many molecules of O2 will be formed from 34.35 grams of KClO3?

b. How many grams of KCl will be formed from 52.53 grams of KClO3?

                     a)  2.53 × 10²³ molecules of O₂

                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

Step 1: Calculate Moles for given amount of KClO₃;

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: Find out  moles of O₂ produced;

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

                     X =  0.42 moles of O₂

Step 3: Calculate No. of Molecules of O₂ as,

No. of Molecules  =  Moles × 6.022 × 10²³

No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

Step 1: Calculate Moles for given amount of KClO₃;

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: Find out  moles of KCl produced;

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: Calculate Mass of KCl as;

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

your answer is c) plasma

~*"auba14"*~