Question

For the balanced equation shown below, how many grams of BaO₂ reacted, if 2.04 moles of BaO are produced?

Remember to use two decimal places and NO units in your answer.
2 BaO₂ → 2 BaO + O₂
Molar masses (g/mol): Ba = 137; O = 16.

Answer 1

To find the mass of barium peroxide (BaO2) that reacted to produce 2.04 moles of barium oxide (BaO), we multiplied the moles of BaO produced by the molar mass of BaO2 (169 g/mol), yielding 344.76 grams of BaO2.

Step-by-step explanation:

In order to calculate the number of grams of barium peroxide (BaO2) that reacted to produce 2.04 moles of barium oxide (BaO), we refer to the balanced chemical equation given: 2 BaO2 → 2 BaO + O2. This indicates a 1:1 molar ratio between BaO2 and BaO. Since molar masses are provided for barium (Ba) and oxygen (O), we can calculate the molar mass of BaO2 as follows:

Molar mass of BaO2 = Ba + 2 × O = 137 + 2(16)

= 169 g/mol

To find out the mass of BaO2 that reacted, we multiply the amount of moles of BaO produced by the molar mass of BaO2:

Mass of BaO2 = 2.04 moles × 169 g/mol

= 344.76 grams

Therefore, 344.76 grams of BaO2 reacted to produce the given amount of BaO.