How many grams of cl2 (mw = 71.0 g/mol) can be prepared from the reaction of 16.0 g of mno2 (fw = 86.9 g/mol) and 30.0 g of hcl (fw = 36.5 g/mol) according to the following chemical equation? mno2 + 4 hcl ® mncl2 + cl2 + 2 h2o0.82 g5.8 g13.0 g14.6 g58.4 g

The mass of chlorine produced in the given reaction is 13.206 grams.

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of manganese oxide = 16.0 g

Molar mass of manganese oxide = 86.9 g/mol

Putting values in equation 1, we get:

Given mass of hydrochloric acid = 30.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of manganese oxide reacts with 3 moles of hydrochloric acid.

So, 0.184 moles of manganese oxide will react with = of hydrochloric acid

As, given amount of hydrochloric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, manganese oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of manganese oxide reacts with 1 moles of chlorine.

So, 0.184 moles of manganese oxide will react with = of chlorine.

Now, calculating the mass of chlorine from equation 1, we get:

Molar mass of chlorine = 71 g/mol

Moles of chlorine = 0.186 moles

Putting values in equation 1, we get:

Hence, the mass of chlorine produced in the given reaction is 13.206 grams.

the answer is   c.) 1575 j.

explanation:

the answer is 0.42 is that is an option. this question provides no info