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Romashka-Z-Leto [24]
3 years ago
5

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
Chemistry
1 answer:
DedPeter [7]3 years ago
7 0
Mass = 473.2 g
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.
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