Answer:
18.8 g
Explanation:
The equation of the reaction is;
AgClO3(aq) + LiBr(aq)------>LiClO3(aq) + AgBr(s)
Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles
Number of moles of LiBr = 10.23 g/86.845 g/mol = 0.1 moles
Since the molar ratio is 1:1, LiBr is the limiting reactant
Molar mass of solid AgBr = 187.77 g/mol
Mass of precipitate formed = 0.1 moles * 187.77 g/mol
Mass of precipitate formed = 18.8 g
Answer:
It's answer is pure substances
Answer:
Explanation:
In sulfur dioxide there are 32.06 grams of sulfur and 32 grams of oxygen .
In sulfur trioxide there are 32.06 grams of sulfur are combined with 48 grams of oxygen.
The ratio of oxygen which reacts with 32.06 gram of sulfur is 32: 48 .
This ratio is equal to 2 : 3.
This is in accordance with law of multiple proportion because , the ratio of mass of oxygen which reacts with constant mass of sulfur is integral ratio . Hence they are in accordance with law of multiple proportions.
<span>For 1 mole of MgCl2, it would require 2 moles of KOH. ( 1 : 2 mole ratio)
Since you have 3 moles of KOH, it is in excess, and MgCl2 is the limiting reactant.</span><span>
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