Question

How many grams of silver bromide are produced when 505 grams of cobalt (III) bromide reacts completely in the following equation:

1 CoBr3 + 3 AgNO3 à 3 AgBr + 1 Co(NO3)3

Answer 1

Mass = 473.2 g
Step-by-step explanation:
Given data:
Mass of cobalt(III) nitrate = 206 g
Mass of silver bromide produced = ?
Solution:
Chemical equation:
CoBr₃ + 3AgNO₃ → 3AgBr + Co(NO₃)₃
Number of moles of cobalt(III) nitrate:
Number of moles = mass/ molar mass
Number of moles = 206 g/ 245 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of cobalt(III) nitrate with silver bromide.