How many grams of water can be formed from the reaction of 8.76 grams of H2 with 10.5 liters of O2 (at STP) according to the balanced equation: 2 H2 + O2 --> 2 H2O Express your answer to 3 sig figs. Do NOT include units!

Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:

Moles H₂ -Molar mass: 2.016g/mol-:

8.76g * (1mol / 2.016g) = 4.345 moles

Moles O₂:

PV = nRT

PV/RT = n

P = 1atm at STP

V = 10.5L

R = 0.082atmL/molK

T = 273.15K at STP

n = 1atm*10.5L / 0.082atmL/molK*273.15K

n = 0.469 moles of oxygen

For a complete reaction of 4.345 moles moles of hydrogen are required:

4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant

Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:

0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.

The mass is -Molar mas H₂O = 18.01g/mol-:

0.938 moles * (18.01g/mol) =