Propane, C3H8 (approximate molar mass = 44 g/mol) is used in gas barbeques and burns according to the thermochemical equation: C3H8(g) + 5 O2(g)3 CO2(g) + 4 H2O(g) AH =-2046 kJ. If it takes 1.8 x 10 kJ to fully cook a rack of lamb on a gas barbeque, how many grams of propane will be required, assuming all the heat from the combustion reaction is absorbed by the lamb? Give your answer to 2 significant figures, but not in scientific notation.

icon
Related questions
Question

please help with the following!

Propane, C3H8 (approximate molar mass = 44 g/mol) is used in gas barbeques and burns according to the thermochemical equation: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4
H20(g) AH =-2046 kJ. If it takes 1.8 x 10° kJ to fully cook a rack of lamb on a gas barbeque, how many grams of propane will be required, assuming all the heat from the
%3D
combustion reaction is absorbed by the lamb? Give your answer to 2 significant figures, but not in scientific notation.
Transcribed Image Text:Propane, C3H8 (approximate molar mass = 44 g/mol) is used in gas barbeques and burns according to the thermochemical equation: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H20(g) AH =-2046 kJ. If it takes 1.8 x 10° kJ to fully cook a rack of lamb on a gas barbeque, how many grams of propane will be required, assuming all the heat from the %3D combustion reaction is absorbed by the lamb? Give your answer to 2 significant figures, but not in scientific notation.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

Blurred answer
Knowledge Booster
Colloids
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.