How much water can be produced from the
complete reaction of 3.64 L of oxygen gas, at STP, with hydrogen gas?
Answer in units of g
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To find the amount of water produced from the complete reaction of oxygen gas with hydrogen gas, we need to know the balanced chemical equation for the reaction. In this case, it is the reaction between hydrogen and oxygen gases to form water:
2H₂ + O₂ → 2H₂O
From the balanced equation, we can see that for every 2 moles of hydrogen gas (H₂) reacted, we get 2 moles of water (H₂O) produced.
To determine the number of moles, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume of gas (3.64 L in this case)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP = 273 K)
Rearranging the equation to solve for n:
n = PV / RT
Substituting the given values:
n = (1 atm) * (3.64 L) / (0.0821 L·atm/(mol·K) * 273 K)
Calculating this, we find the number of moles of oxygen gas (O₂) reacted.
Next, we can use the balanced equation to determine the number of moles of water (H₂O) produced. Since the balanced equation shows that 2 moles of hydrogen (H₂) react to form 2 moles of water (H₂O), the number of moles of water produced would be the same as the number of moles of hydrogen.
Finally, to convert the number of moles of water (H₂O) to grams, we can use the molar mass of water, which is approximately 18.015 g/mol.
Multiply the number of moles of water (H₂O) with the molar mass of water to find the mass in grams.
This calculation will give you the answer in units of g.