Silver bromide is produced from 35.5 g of silver nitrate in the following reaction:? 2AgNO3(aq) + MgBr2(aq) → 2AgBr(s) + Mg(NO3)2(aq)?
(a) How many moles of silver nitrate are present?
(b) How many moles of magnesium nitrate will produced?
(c) Find the mass of magnesium nitrate produced.
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(a)
(35.5 g AgNO3) / (169.87323 g AgNO3/mol) = 0.20898 = 0.209 mol AgNO3
(b)
Supposing the MgBr2 to be in excess:
(0.20898 mol AgNO3) x (1 mol Mg(NO3)2 / 2 mol AgNO3) = 0.10449 mol = 0.104 mol Mg(NO3)2
(c)
(0.10449 mol Mg(NO3)2) x (148.3148 g Mg(NO3)2/mol) = 15.5 g Mg(NO3)2