When thirty mL of a 0.2000 M solution of sodium oxalate (Na2C2O4) is added to 25

Question

When thirty mL of a 0.2000 M solution of sodium oxalate (Na2C2O4) is added to 25.00 mL of a 0.1500 M solution of lanthanum (III) chloride, lanthanum oxalate precipitates. The balanced equation for the reaction is a.) How many grams of La(C2O4)3 (s) are obtained, assuming 100% yield? b.) What are the concentrations (M) of La3+, C2O4^2-, Na+, and Cl- after the reaction is complete? Assume that volumes are additive. 5. Nitrogen dioxide gas reacts with water vapor in the atmosphere to produce acid rain (hydrogen nitrate is and nitrogen monoxide gas). a.) Write a balanced equation for this chemical reaction. b.) Assume that the temperature and pressure remain constant. How many liters of hydrogen nitrate gas at 50 torr and 40.0degree C are produced when 560 g of nitrogen dioxide are allowed to react? 7.(15-pts.) Vessel A contains 4.0 liters of CO at a pressure of 1.5 atm. Vessel 13 has a volume of 3.0 liters id contains Ar gas at 6.2 atm. The volume of vessel C is 8.0 liters and it contains Xe at a pressure of 4.4 atm. Assume that the temperature for the whole system remains constant at 25.0 degreeC. a.) What is the predicted pressure after stopcock 2 is opened and the Ar and Xe are mixed? Assume that volume of the stopcock and connecting tubing is negligible. b.) What is the predicted pressure if both stopcocks are opened, allowing the three gases to mix together.

Explanation / Answer

a) if stopcock 2 is opened

Volume of the mixer = 3+8 = 11 L

T = constant = 298 k

No of moles of B(Ar) = PV/RT = 6.2*3/(0.0821*298) = 0.76 mol

No of mole sof C(Xe) = 4.4*8/(0.0821*298) = 1.44 mol

No of moles of mixer = 1.44+0.76 = 2.2 mol

Pressure of mixer after opening stopcock2 = nRT/V = 2.2*0.0821*298/11

    = 4.9 atm

b) if both stopcocks opened.

No of moles of A(CO) = 1.5*4/(0.0821*298) = 0.24 mol

Volume of the mixer = 3+8+4 = 15 L

No of moles of mixer = 1.44+0.76+0.24 = 2.44 mol

T = constant = 298 k

Pressure of mixer after opening stopcock1,2 = nRT/V

= 2.44*0.0821*298/15

= 3.98 atm

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