In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g)

The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 > 8CO2 + 6H2O 

number of moles = mass / molar mass 
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams

0.54 g
       
Lookup the atomic weights of oxygen and hydrogen. Atomic weight oxygen = 15.999 Atomic weight hydrogen = 1.00794 Calculate molar masses Molar mass O2 = 2 * 15.999 = 31.998 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now determine how many moles of O2 you have 2.1 g / 31.998 g/mol = 0.065629102 mol Balance the equation so you get as the balanced equation, 2 C4H6(g) + 11 O2(g) → 8 CO2(g) + 6 H2O(g) Looking at the balanced equation, for every 11 moles of O2 consumed, 5 moles of H2O are produced. So divide the number of moles of O2 you have by 11, then multiply by 5, giving: 0.065629102 mol / 11 * 5 = 0.02983141 mol And finally, multiply by the molar mass of H2O, so 0.02983141 mol * 18.01488 g/mol = 0.53740927 g Round the results to 2 significant figures getting 0.54 g

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