If 34.0 grams of H2 reacted with 23.8 grams of O2 in order to make water, how many grams of water would be produced? Please help! I dont understand how to do this. Please show work.
2H2 + O2 ---> 2H2O
2H2 + O2 ---> 2H2O
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Calculate the number of moles of each reactant:
# mol of H2 = 34.0g/2.02gmol-1 = 16.83 mol
#mol of O2 = 23.8g/32.00gmol-1 = 0.744 mol
Therefore, you will run out of oxygen first (it is the limiting reagent)
In the equation, one mole of oxygen produces two moles of water. Therefore, 0.74 moles of oxygen will produce 1.488 moles of water.
The mass of 1.488 moles of water is: 1.488 mol x 18.02 g mol-1 = 26.8g
# mol of H2 = 34.0g/2.02gmol-1 = 16.83 mol
#mol of O2 = 23.8g/32.00gmol-1 = 0.744 mol
Therefore, you will run out of oxygen first (it is the limiting reagent)
In the equation, one mole of oxygen produces two moles of water. Therefore, 0.74 moles of oxygen will produce 1.488 moles of water.
The mass of 1.488 moles of water is: 1.488 mol x 18.02 g mol-1 = 26.8g