Question

Assignment Score: 233/2000 Resources Hint Check Answer Question 19 of 20 > A sample of 5.17 g of liquid 1-propanol, C,H,O, is combusted with 46.5 g of oxygen gas. Carbon dioxide and the products. Write the balanced chemical equation for the reaction. Physical states are optional. chemical reaction: What is the limiting reactant? O oxygen O l-propanol How many grams of Co, are released in the reaction? mass of CO, How many grams of the excess reactant remain after the reaction is complete? about us Careers privacy policy Terms of use contact us hele

Answer 1

60.09 g/mol 32.00 g/mol Molar mass of C3H7OH = Molar mass of O2 44.01 g/mol Molar mass of CO2 18.02 g/mol Molar mass of H2O Balanced chemical equation: 2 C3H7OH 9 02 6 CO2 8 H20 Theoretical moles ratio of C3H7OH 2 9 02 Calculating number of moles: Mass of C3H7ОН %3 5.17 g Moles Mass /Molar mass Moles of C3H7OH 60.09 5.17 Moles of C3H7OH 0.086 mol 46.5 g Mass of O2 Moles= Mass/ Molar mass = ( 46.5 Moles of O2 32.00 Moles of 02 1.453 mol Experimental moles ratio of C3H7OH 0.086 1.453 O2 1.000 16.889 2.000 33.779 Limiting reactant is C3H7OH 1-propanol Excess reactant is O2 II

Mass of products depends upon limiting reactant From balanced equation: 2 moles of produces 6 moles of CO2 СЗН7ОH 0.086 moles of СЗн7оH produces? moles of CO2 2) x 0.086 6 moles of CO2 moles of CO2 0.258 mol Calculating mass: Moles x Molar mass Mass CO2 0.258 x 44.01 mass of 11.360 g CO2 mass of Excess reactant is 02 From balanced equation: 9 moles of O2 2 moles of СЗН7ОH reacts with moles of 02 0.086 moles of СЗн7оH reacts with ? = (9 / 2 )x 0.086 moles of 02 0.387 mol moles of O2 0.387 1.066 mol Moles of excess reactant remain = 1.453 Calculating mass: Mass Moles x Molar mass x 32.00 34.11 1.066 O2 mass of g 34.11 g mass of excess reactant remain =