6.03 Calorimetry Lab Report

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The first step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to balance the equation like this 2 C2H6 + 7 O2 ----> 4 CO2 + 6 H20. The second step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to find the mole ratio which is 2 moles of C2H6 : 6 moles of H20. The third step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to multiply this by the mole ratio like this 6 moles of water / 2 moles of ethane * 11.7 moles of ethane = 35.1 moles of water. The fourth step that I did to find the number of grams

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