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Limiting Reactant Lab Report

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The first step that I did was to set up a equation with a chemical reaction like this 2Mg + O2 = 2MgO. The second step that I did was to find out which of the reactants are limiting by multiplying the number of moles of the reactant to the coefficient that is stoichiometric. The third step is to find how many moles magnesium has by forming the equation 97.2 grams * moles / 24 grams * 2 and got the answer of 8 moles The fourth step that I did was to find how many moles oxygen has by creating the equation 88.5 grams * moles / 32 grams * 1 and got the answer of 2.7 moles. Now I have discovered that oxygen has the smaller amount of moles, this is the limiting reactant. The last step that I did was to solve this problem that is based on the number of oxygen being used. …show more content…
The first step that I did to find the limiting reactant was to balance the equation out like this 4Li + O2 ----> 2Li2O. The second step that I did to find the limiting reactant was to divide the number of moles to its matching coefficient that is stoichiometric. The third step that I did was to figure out the limiting reactant by using the numbers that are associated in front of each of the reactants as a ratio like this 1:2 for O2 to Li2O and 4:2 ratio for Li to Li20. The fourth step that I did was to solve the problem by using the equation 8.4 / 4 = 2.1 for lithium and for oxygen the result is 4.6, concluding that the limiting reactant when 8.4 moles of lithium reacts with 4.6 moles of oxygen gas is lithium. The first step that I did for this problem was to balance out this equation like this 2C6H6 + 15O2 = 6H2O + 12CO2. The second step that I did was to divide the number of moles to its matching coefficient that is stoichiometric like this n(C6H6)/2 =

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