WILL MARK U BRAINLIEST AND FOLLOW YOU AND WILL SAY THANKS FOR ALL YOUR ANSWER HOW MANY GRAMS OF AgBr CAN BE PRODUCED FROM 50g OF MgBr2 AND 100g OF AgNO3 MgBr2+2AgNO3->2AgBr + Mg(NO3)2
Answers
Explanation:
In a double replacement reaction between silver nitrate and magnesium bromide. Calculate the mass of silver bromide which should be produced from 22.5g of silver nitrate. The actual yield was 22.0g, how do you calculate percent yield?
Chemistry Percent Yield
1 Answer
Stefan V.
Oct 29, 2015
Theoretical yield:
24.9 g
Percent yield:
88.4%
Explanation:
Start by writing the balanced chemical equation for this double replacement reaction
2
AgNO
3(aq]
+
MgBr
2(aq]
→
2
AgBr
(s]
⏐
⏐
↓
+
Mg
(
NO
3
)
2(aq]
Notice that you have a
1
:
2
mole ratio between silver nitrate and silver bromide. This means that the reaction will produce
1
mole of the latter for every
1
moel of the former that takes part in the reaction.
However, keep in mind that this is true for a reaction that has a
100
%
yield - this represents the theoretical yield of the reaction.
If the percent yield of the reaction is smaller than
100
%
, than silver bromide will not be produced in a
1
:
1
mole ratio with silver nitrate.
So, how many moles of silver nitrate do you have in
22,5 g
of silver nitrate? Use the compound's molar mass to get
22.5
g
⋅
1 mole AgNO
3
169.87
g
=
0.13245 moles AgNO
3
So, what would the theoretical yield of the reaction be?
Well, the
1
:
1
mole ratio would produce
0.13245
moles AgNO
3
⋅
1 mole AgBr
1
mole AgNO
3
=
0.13245 moles AgBr
To calculate how many grams of silver bromide would contain this many moles? Once again, use the compound's molar mass
0.13245
moles
⋅
187.77 g
1
mole
=
24.9 g AgBr
Now, a reaction's percent yield is defined as
%
yield
=
actual yield
theoretical yield
×
100
If the actual yield is
22.0 g
, then the percent yield of the reaction is
% yield
=
22.0
g
24.9
g
×
100
=
88.4%
Here's a photo of how the silver bromide precipitate would