How much water can be produced from the

complete reaction of 2.56 L of oxygen gas, at
STP, with hydrogen gas
answer in units of g.

Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the amount of water that can be produced from the complete reaction of oxygen gas with hydrogen gas, we first need to write the balanced equation for the reaction:

2 H₂ + O₂ → 2 H₂O

From the balanced equation, we can see that for every mole of oxygen gas (O₂), we will produce 2 moles of water (H₂O).

To find the moles of oxygen gas (O₂) in 2.56 L at standard temperature and pressure (STP), we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure = 1 atm (STP)
V = volume = 2.56 L
n = moles
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature = 273 K (STP)

Rearranging the equation to solve for moles (n):

n = PV / RT

Substituting the given values:

n = (1 atm) * (2.56 L) / (0.0821 L·atm/(mol·K) * 273 K)

n ≈ 0.1275 moles

Since the stoichiometry of the reaction tells us that 2 moles of water (H₂O) are produced for every mole of oxygen gas (O₂), we can calculate the moles of water produced:

moles of water = 2 * 0.1275 = 0.255 moles

To convert moles of water (H₂O) to grams, we need to use the molar mass of water, which is approximately 18 g/mol.

mass of water = moles of water * molar mass of water

mass of water ≈ 0.255 moles * 18 g/mol

mass of water ≈ 4.59 g

Therefore, the amount of water that can be produced from the complete reaction of 2.56 L of oxygen gas, at STP, with hydrogen gas is approximately 4.59 grams.