Writing And Balancing Chemical Reactions, Stoichiometry, Limiting Reactants Worksheet With Answers - Chem110 Page 8
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8CHEM110 Week 5 Notes (Stoichiometry)
Page 8 of 8
O
remains in excess, start with the limiting reactant, Al, and calculate the mass of O
that
2
2
would be needed.
5.0 g Al
1 mol Al
3 mol O
31.998 g O
2
2
26.982 g Al
4 mol Al
1 mol O
2
= 4.4g O
needed (2SD). The mass of O2 that remains is 5.0 - 4.4 = 0.6g O
remains.
2
2
Here are the results from above in a BCA table:
4Al(s) + 3O
(g) → 2Al
O
(s)
2
2
3
B, g
5.0
5.0
0
B, mol 0.185 0.156
0
C, mol -0.185 -0.139
+0.0927
A, mol 0
0.017
0.0927
A, g
0
0.6
9.4
Chemical Analysis
Stoichiometry can be used to analyze samples for percent composition by mass.
Chalcanthite is an impure mineral containing CuSO
O. Determine the percentage of
⋄5H
4
2
copper(II) sulfate pentahydrate in the mineral from these data. A 5.13g sample of
chalcanthite is crushed, and the copper(II) sulfate is dissolved in water. The blue copper
(II) sulfate solution is filtered and then precipitated as 0.91 copper(II) phosphate using
sodium phosphate.
Step 1: Write the balanced chemical reaction.
Step 2: Convert the 0.91g of copper(II) phosphate to moles.
Step 3: Convert moles of copper(II) phosphate to moles of copper(II) sulfate.
Step 4: Convert moles of copper(II) sulfate to moles of copper(II) sulfate pentahydrate.
Step 5: Convert moles of copper(II) sulfate pentahydrate to grams.
Step 6: Determine the precent of copper(II) sulfate pentahydrate in the mineral.
Step 1: 3CuSO
(aq) + 2Na
PO
(aq) → 3Na
SO
(aq) + Cu
(PO
)
(s)
4
3
4
2
4
3
4
2
Steps 2-5:
0.91 g Cu
(PO
)
1 mol Cu
(PO
)
3 mol CuSO
1 mol CuSO
O
249.683 g CuSO4⋄5H2O
⋄5H
3
4
2
3
4
2
4
4
2
380.578 g Cu
(PO
)
1 mol Cu
(PO
)
1 mol CuSO
1 mol CuSO
O
⋄5H
3
4
2
3
4
2
4
4
2
=1.79g CuSO
O (2SD)
⋄5H
4
2
Step 6: 1.79g CuSO
O / 5.13g sample * 100% = 35% CuSO
O (2SD)
⋄5H
⋄5H
4
2
4
2
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