Question

NEED ANSWERS ASAP PLZ

QUESTION 15 9.00 grams of C2H4 is reacted with 5.50 gram F2 according to the following equation. How many gram of HF can be formed, assuming 100% yield? (i.e., what is theoretical yield?) C2H4 + 6F2 --> 2CF4 + 4 HF Hint: Find limiting reagent. O 0.281 g 0 0.830 g O 1.93 g O 17.219 0 20.55 g O 19.34 g 0 12.04 g 0 25.91 g

Answer 1

Molar mass of C2H4 = 28.05 g/mol Molar mass of F2 = 38.00 g/mol Molar mass of CF4 = 88 g/mol Molar mass of HF = 20.01 g/mol 1 : 6 Balanced chemical equation: 1 C2H4 + 6 F2 + 2 CF4 + 4 HF Theoretical moles ratio of C2H4 : F2 = Calculating number of moles: Mass of C2H4 = 9.00 g Moles = Mass / Molar mass Moles of C2H4 = ( 9.00 / 28.1 ) Moles of C2H4 = 0.321 mol Mass of F2 = 5.5 g Moles = Mass / Molar mass Moles of F2 = ( 5.50 / 38.0 ) Moles of F2 = 0.145 mol Experimental moles ratio of C2H4 : F2 = = 0.321 1.000 : : 0.145 0.451 Limiting reactant is F2 Excess reactant is C2H4 Mass of products depends upon limiting reactant From balanced equation: 6 moles of F2 produces 4 moles of HF 0.145 moles of F2 produces ? moles of HF moles of HF = ( 4 / 6 ) * 0.145 moles of HF = 0.097 mol Calculating mass: Mass = Moles Molar mass mass of HF = 0.097 * 20.01 mass of HF = 1.93 g