I did a calculation but the answer is wrong. Please help. Thank you Molar mass S

Question

I did a calculation but the answer is wrong. Please help. Thank you

Molar mass SO3 = 80.1 g/mol

Moles of SO3 = 1.17g/ (80.1g/mol) = 0.0146 moles

P(SO3)= nRT/V = (0.0146 × 0.08206 × 700)/2 = 0.4193 atm

Molar mass of O2 =32g /mol

Moles of O2 = 0.107g/(32g/mol) = 0.0033 moles

P(O2) = nRT/V =( 0.0033×0.08206 × 700)/2 = 0.0947 atm

Kp = 3 × 104 = PSO32 /( PO2 × PSO22) = (0.4193)2/( 0.0947 × PSO22 )

Pso2 )2= 0.1758 /(0.0947 × 30000) = 6.188 × 10-5

Pso2 = 0.00786 atm

n = PV/RT = 0.000273 moles

0.000273 moles /(64.1 g/ mol) = 0.0000042grams

Part A For 2 SO2 (g) + O2 (g) 2S03 (g), K, 3.0 × 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.17 g of SO3 and 0.107 g of O2 How many grams of S02 are in the vessel? Express your answer using two significant figures. 0 m= Submit equest Answer Provide Feedback

Explanation / Answer

you did everything right except last step.

i will correct that for you

you got moles of SO2 = 0.000273

now weight of SO2 = moles * molecular weight

= 0.000273 * 64.1 (Mol. weight of SO2 = 64.1gm/mol)

so weight of SO2 = 0.0174993 gram

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.