Answer:
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.
Co(NO₃)₃ : AgBr
1 : 3
0.84 : 3/1 × 0.84 = 2.52 mol
Mass of silver bromide:
Mass = number of moles × molar mass
Mass = 2.52 mol × 187.77 g/mol
Mass = 473.2 g
The mass of silver bromide produced is 473.744 grams
The chemical equation for the reaction between cobalt (III) bromide and silver nitrate produces silver bromide and cobalt (III) bromide.
i.e.
[tex]\mathbf{COBr_3 + 3Ag(NO_3) \to 3AgBr + Co(NO_3)_3}[/tex]
Number of moles = mass/molar mass
Number of moles = 206 gram /244.95 g/mol
Number of moles = 0.8410 moles
From the above chemical reaction, if 3 moles of AgBr and 1 mole of Co(NO)₃ is produced.
Then, 0.8410 moles of Co(NO)₃ will be equivalent to (3 × 0.8410) moles of AgBr
∴
Number of moles of AgBr = 2.523 moles
The molar mass of AgBr = 187.77 g/mol
Number of moles = mass/ molar mass
Mass of AgBr = number of moles of AgBr × Molar mass of AgBr
Mass of AgBr = 2.523 moles × 187.77 g/mol
Mass of AgBr = 473.744 grams
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