Please help! Initial: CaCl2•2H2O (g) 1.0 Initial: CaCl2•2H2O (moles) 0.068 Initi
ID: 494023 • Letter: P
Question
Please help!
Initial: CaCl2•2H2O (g)
1.0
Initial: CaCl2•2H2O (moles)
0.068
Initial: CaCl2 (moles)
0.068
Initial: Na2CO3 (moles)
0.068
Initial: Na2CO3 (g)
.7209
Theoretical: CaCO3 (g)
0.680
Mass of Filter paper (g)
1
Mass of Filter Paper + CaCO3 (g)
1.7
Actual: CaCO3 (g)
.72
% Yield:
103%
show work for all calculations required to generate the data in all the data tables from the lab. Show work for percent error calculation.
Conclusion: Summarize the main experimental finding of the lab. Calculate percent error for this value.
Initial: CaCl2•2H2O (g)
1.0
Initial: CaCl2•2H2O (moles)
0.068
Initial: CaCl2 (moles)
0.068
Initial: Na2CO3 (moles)
0.068
Initial: Na2CO3 (g)
.7209
Theoretical: CaCO3 (g)
0.680
Mass of Filter paper (g)
1
Mass of Filter Paper + CaCO3 (g)
1.7
Actual: CaCO3 (g)
.72
% Yield:
103%
Explanation / Answer
Molar mass of CaCl2.H2O = 128.98g/mole
weight of CaCl2.H2O = 1.0g
no of moles CaCl2.H2O = 1.0/128.98=0.0078 mole
no of moles of CaCl2 = 0.0068 mole
Initial mole of Na2CO3 = 0.0068mole
Molar mass of Na2CO3 = 106 g/mole
Weight of Na2CO3 = 0.0068 × 106 = 0.7209
Limiting is Na2CO3 and CaCl2.H2O is excess
Na2CO3 + CaCl2.H2O --------> 2NaCl + CaCO3 + H2O
1 mole of Na2CO3 give 1 mole of CaCO3
0.0068 mole of Na2CO3 give 0.0068 mole of CaCO3
Molar mass of CaCO3 = 40.08g/mole
mass of CaCO3 = 0.0068 × 40.08 = 0.273g
Theoretical yield = 0.273g
Actual mass = 0.72g
Actual yield = 0.72g
Percentage yield =( 0.72/0.273)×100 = 263.74
percentage error = ((0.72-0.273)/0.273)×100 = 163.74