Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
Learn more about Partial pressure here:
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