Question

Practice with Stoichiometry Calculations - Problem Solving Workflow: You may be given either the number of grams or moles of chemical A... ...and asked to determine either the number of moles or grams of chemical B Grams of A Moles of A Moles of B Grams of B 1. Convert from the number of 2. Convert from the number grams of A to the number of moles of moles of A to the number of moles of B using the coefficients of the balanced chemical equation 3. Convert from the number of moles of B to the number of grams of B using the molar mass of B. 8. How many grams of Hz are required to produce 7.50 grams of ammonia NH,? Use the balanced equation (this time the equation is already balanced): N2(g) + 3H2(g) → 2NH3(g) 9. If 54.7 grams of propane (CH) and 89.6 grams of oxygen (O2) are available in the balanced combustion reaction below: C3Hg +502 3C02 + 4H20 a) Which reactant is limiting (show your work)? b) What is the theoretical yield of CO2 in grams?

Answer 1

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3CO2 (g)4 H20 (g) The balanced equation is C3H8 (g) + 5 02 (g) Сзн8 gas Mass of 54.700 Molar mass of CO2 44.01 g mol-1 g Mol wt of C3H8 gas g mol-1 Mass/Molar Mass 44.10 Moles of C3H8 gas 54,700 44.1 g mol-1 + g mol 1.2404 Mass of 89.600 O2 gas ge Mol wt of 02 gas 32.000 gmol-1 Mass/Molar Mass Moles of 02 gas 89.600 32 g mol-1 + 2.8000 mol C3H8 gas (3 mole H20 1 mol C3H8) (3 mole H20/1 mol сзн8) Сзн8 gas Moles of from Moles of CO2 mol x 1.2404 3.7211 mol Moles of C3H8 gas (3 mole CO 2/5 mol 02) from 02 gas Moles of CO2 (3 mole Co 2/5 mol 02) 2.8000 mol x mol 1.6800 produce less moles of product. So is limiting reagent and C3H8 gas is excess reagent 02 gas 02 gas So Theoretical yiled of CO2 Moles x Molar Mass 1.6800 mol x g mol-1 44.010 73.937 g