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Clearly, four moles of water result from each mole of propane.
And thus .....
How does the total mass of the products compare to the total mass of the reactants? Is this significant conceptually?
( a) In the complete combustion of propane, how many moles of H2O(l) are produced per mole of CO2(g)? (b) A 1.638-g sample of propane is burned in excess oxygen. What are the theoretical yields (in grams) of CO2(g) and H2O(l) expected from the reaction
How molecules of CO2 (g) are produced when 0.75 moles of propane are burned? Chemical reaction: C3H8 + 5O2 à 3CO2 + 4H2O
The combustion of propane (C3H8) produces CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) The reaction of 7.5 mol of O2 with 1.4 mol of C3H8 will produce ________ mol of CO2. Group of answer choices.
Consider the reaction below. If you start with 7.0 moles of C3H8 (propane) and 7.0 moles of O2, _ is the percent yield if 4.0 moles of carbon dioxide is produced. C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
Consider the combustion of propane: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l) ΔH = –2221 kJ Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.
How many grams of O2(g) are needed to completely burn 35.9 g of C3H8(g)? C3H8 + 5O2 ---> 3CO2 + 4H2O
The combustion of propane, C3H8, occurs via the reaction C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C3H8 (g)= -104.7 CO2(g)= −393.5 H2O(g)= −241.8 Calculate the enthalpy for the combustion of 1 mole of propane.
Given the following equation, C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (g); H comb = -2219 kJ/mole A) how much heat will be released when 10 g of propane is burned? B) how much heat will be released when 10 g of water (H2O) is formed? Please show all work.
Please help? Calculate deltaHr for the reaction: C3H8(g) + 5O2(g)----->3CO2(g) + 4H2O(l) Given: 3C(s) + 4H2(g)----->C3H8(g) deltaH -24.8 kcal H2(g) + 1/2 O2(g)----->H2O(l) deltaH -68.3kcal C(s) + O2(g)-------> CO2(g) deltaH -94.0 kcal
GIVEN INFO LP gas burns according to the following exothermic reaction: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) ΔH∘rxn= −2044kJ What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 C) to boiling (100.0 C)? Assume that during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings. Express your answer using two significant figures. FINAL ANWSER SHOULD BE IN GRAMS ?