Question

5) If 23, of CyHie Were Combusted with 31q 0f liaiting How much of the non-latiag feageat willbe leet ouer aekr thereacton is Comple ted ? one CeagentS O4gen

Answer 1

58.12 g/mol Molar mass of C4H10 32.00 g/mol Molar mass of O2 Balanced chemical equation: 10 H20 13 O2 8 CO2 2 C4H10 Theoretical moles ratio of 2 13 C4H10 02 : Calculating number of moles: Mass of C4H10 23 Moles Mass / Molar mass /58.12 Moles of C4H10 - ( 23 Moles of C4H10 0.396 mol O2 37 Mass of g Moles = Mass / Molar mass /32.00 Moles of 37 O2 1.156 mol Moles of O2 0.396 1.156 Experimental moles ratio of C4H10 O2 1.000 2.922 2.000 5.844 II

Limiting reactant is O2 Excess reactant is C4H10 Mass of products depends upon limiting reactant From balanced equation: 13 moles of 02 reacts with 2 moles of C4H10 1.156 moles of 02 reacts with ? moles of C4H10 2 13 x 1.156 moles of C4H10 moles of C4H10 0.178 mol Moles of excess reactant remain 0.178 0.218 mol 0.396 Calculating mass: Moles x Molar mass Mass 12.66 g C4H10 0.218 x 58.12 mass of 12.66 g mass of excess reactant remain