<span>Answer:
Zn(2+) + 2e(-) -------> Zn
1 mole of Zn is deposited by 2F of electricity ...
so 48.9 mole of Zn will be deposited by 48.9 X 2F = 97.8 F of electricity...
as 1F = 96500 C
so 97.8 F = 97.8 X 96500 = 9437700 C of electricity...</span>
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
The number of grams in 1.70 moles of Ca(NO₃)₂ is 384.2 grams
<h3>How to determine the mass of Ca(NO₃)₂</h3>
The mole of a substance is related to it's mass and molar mass according to the following equation:
Mole = mass / molar mass
With the above formula, we can determine the mass of Ca(NO₃)₂ as illustrated below:
- Mole of Ca(NO₃)₂ = 1.70 moles
- Molar mass of Ca(NO₃)₂ = 40 + 3[14 + (16 × 3)] = 40 + 3[14 + 48] = 40 + 3(62) = 40 + 186 = 226 g/mol
- Mass of Ca(NO₃)₂ = ?
Mole = mass / molar mass
1.70 = Mass of Ca(NO₃)₂ / 226
Cross multiply
Mass of Ca(NO₃)₂ = 1.70 × 226
Mass of Ca(NO₃)₂ = 384.2 grams
Thus, the mass of 1.70 moles of Ca(NO₃)₂ is 384.2 grams
Learn more about mole:
brainly.com/question/13314627
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Your answer would have to be #3
