Question

For tl e following reaction, 5.10 grams of propane (C3Hg) are allowed to react with 21.8 grams of oxygen gas. prop ne (CHg) (g) + oxygen (g) carbon dioxide (g) + water (g) What is the maximum amount of carbon dioxide that can be formed? grams Wha is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

1 : 5 Molar mass of C3H8 = 44.1 g/mol Molar mass of O2 = 32.00 g/mol Molar mass of CO2 = 44.01 g/mol Molar mass of H20 = 18.02 g/mol Balanced chemical equation: 1 C3H8 + 5 O2 + 3 CO2 + 4 H20 Theoretical moles ratio of C3H8 : 02 = Calculating number of moles: Mass of C3H8 = 5.1 g Moles = Mass / Molar mass Moles of C3H8 =( 5.1 / 44.10 ) Moles of C3H8 = 0.116 mol Mass of 02 = 21.8 g Moles = Mass / Molar mass Moles of 02 = ( 21.8 / 32.00) Moles of 02 = 0.681 mol Experimental moles ratio of C3H8 : 02 = = 0.116 : 0.681 1.000 : 5.891

Limiting reactant is C3H8 propane Excess reactant is 02 Mass of products depends upon limiting reactant From balanced equation: 1 moles of C3H8 produces 3 moles of CO2 0.116 moles of C3H8 produces ? moles of CO2 moles of CO2 = ( 3 / 1 )* 0.116 moles of CO2 = 0.347 mol Calculating mass: Mass = Moles x Molar mass mass of CO2 = 0.347 x 44.01 mass of CO2 = 15.269 g Excess reactant is 02 From balanced equation: 1 moles of C3H8 reacts with 5 moles of 02 0.116 moles of C3H8 reacts with ? moles of 02 moles of O2 = ( 5 / 1 ) x 0.116 moles of O2 = 0.578 mol Moles of excess reactant remain = 0.681 - 0.578 = 0.103 mol Calculating mass: Mass = Moles x Molar mass mass of 2 = 0.103 x 32.00 = 3.30 g mass of excess reactant remain = 3.30 g