haileyparrill703
05.08.2019 •
Chemistry
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
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Ответ:
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)
For manganese oxide:Given mass of manganese oxide = 16.0 g
Molar mass of manganese oxide = 86.9 g/mol
Putting values in equation 1, we get:
For hydrochloric acid: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.
Ответ:
Step-by-step explanation: