Although many sulfate salts are soluble in water, calcium sulfate is not (Table 7. 1). Therefore, a solution of calcium chloride will react with sodium sulfate solution to produce a precipitate of calcium sulfate. The balanced equation is :math> CaCl 2 ( a q ) + Na 2 SO 4 ( a q ) → CaSO 4 ( s ) + 2 NaCl ( a q ) a solution containing 5.21 g of calcium chloride is combined with a solution containing 4.95 g of sodium sulfate, which is the limiting reactant? Which reactant is present in excess?
Although many sulfate salts are soluble in water, calcium sulfate is not (Table 7. 1). Therefore, a solution of calcium chloride will react with sodium sulfate solution to produce a precipitate of calcium sulfate. The balanced equation is :math> CaCl 2 ( a q ) + Na 2 SO 4 ( a q ) → CaSO 4 ( s ) + 2 NaCl ( a q ) a solution containing 5.21 g of calcium chloride is combined with a solution containing 4.95 g of sodium sulfate, which is the limiting reactant? Which reactant is present in excess?
Solution Summary: The author explains that the expected or theoretical yield of a product depends on the limiting reactant of the reaction.
Although many sulfate salts are soluble in water, calcium sulfate is not (Table 7. 1). Therefore, a solution of calcium chloride will react with sodium sulfate solution to produce a precipitate of calcium sulfate. The balanced equation is
:math>
CaCl
2
(
a
q
)
+
Na
2
SO
4
(
a
q
)
→
CaSO
4
(
s
)
+
2
NaCl
(
a
q
)
a solution containing 5.21 g of calcium chloride is combined with a solution containing 4.95 g of sodium sulfate, which is the limiting reactant? Which reactant is present in excess?
When two solutions, one of 0.1 M KCl (I) and the other of 0.1 M MCl (II), are brought into contact by a membrane. The cation M cannot cross the membrane. At equilibrium, x moles of K+ will have passed from solution (I) to (II). To maintain the neutrality of the two solutions, x moles of Cl- will also have to pass from I to II. Explain this equality: (0.1 - x)/x = (0.1 + x)/(0.1 - x)
Calculate the variation in the potential of the Pt/MnO4-, Mn2+ pair with pH, indicating the value of the standard potential. Data: E0 = 1.12.
Given the cell: Pt l H2(g) l dis X:KCl (sat) l Hg2Cl2(s) l Hg l Pt. Calculate the emf of the cell as a function of pH.
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Calorimetry Concept, Examples and Thermochemistry | How to Pass Chemistry; Author: Melissa Maribel;https://www.youtube.com/watch?v=nSh29lUGj00;License: Standard YouTube License, CC-BY