Final answer:
The reaction between cobalt (III) nitrate and silver bromide produces 294 grams of silver bromide when 206 grams of cobalt (III) nitrate are used.
Step-by-step explanation:
The balanced chemical equation for the reaction between cobalt (III) nitrate and silver bromide is:
3CoBr3(aq) + 3AgNO3(aq) → 3AgBr(s) + Co(NO3)3(aq)
According to the stoichiometry of the reaction, for every 3 moles of cobalt (III) nitrate, 3 moles of silver bromide are produced. Therefore, we can create the following ratio:
(3 mol AgBr)/(3 mol CoBr3) = 1 mol AgBr/1 mol CoBr3
To find the grams of silver bromide produced, we need to convert the grams of cobalt (III) nitrate to moles using its molar mass and then use the molar ratio to find the grams of silver bromide. Here's the calculation:
206 g Co(NO3)3 × (1 mol Co(NO3)3/241.11 g Co(NO3)3) × (1 mol AgBr/1 mol Co(NO3)3) × (187.78 g AgBr/1 mol AgBr) = 294 g AgBr
Therefore, option c. 294 g of silver bromide are produced when 206 grams of cobalt (III) nitrate undergo the reaction.