4 moles
also 2 moles of oxygen will be consumed
Balanced equation. 2H2 + O2 -> 2H2O 355 grams O2/32 grams = 11.1 moles O2 check for limiting reactant 11.1 moles O2 (2 mole H2/1 mole O2) = 22.2 mole H2 and H2 has no where near that many moles, so limits and drives reaction so, as they are one to one...... 22.2 moles of H2O are produced
This reaction is:2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2From 4 moles of Al 6 moles of hydrogen are obtained.
CO2 + H2 -> CO + H2O one to one here 30.6 moles H2O (1 mole H2/1 mole H2O) = 30.6 moles Hydrogen gas needed
Oxygen limits the reaction, so......Balanced equation. 2H2 + O2 -> 2H2O 7.89 mole H2O (1 mole O2/2 mole H2O) = 3.95 mole oxygen gas needed ------------------------------------------
Balanced equation first.2H2 + O2 --> 2H2OGet moles products.4 grams H2 (1 mole H2/2.016 grams) = 1.984 moles H264 grams O2 (1 mole O2/32 grams) = 2.000 moles O2I suspect hydrogen gas of limiting and driving the reaction.1.984 moles H2 (1 mole O2/2 moles H2) = 0.992 moles O2 ( you have more than this in equation )2.000 moles O2 (2 mole H2/1 mole O2) = 4.000 moles H2 ( you do not have this much and H2 will drive this reaction )1.984 moles H2 (2 moles H2O/2 moles H2)(18.016 grams/1 mole H2O)= 36 grams water produced====================
2H2 + O2 --> 2H2OFor every 2 moles of H2, 2 moles of H2O will be produced (i.e., a 1:1 ratio). So to produce 8.25 moles of H2O you will also need 8.25 moles of H2
For the reaction 2 H2 + O2 -> 2 H2O, we see that 1 mole of O2 produces 2 moles of H2O. Therefore, to produce 10.2 moles of H2O, we would need 5.1 moles of O2.
To determine the limiting reactant, first convert the masses of N2 and H2 to moles. Then divide the amount of each reactant by its stoichiometric coefficient in the balanced equation to find moles of NH3 produced. The limiting reactant is the one that produces the least amount of NH3.
Balanced equation. 2H2 + O2 -> 2H2O 355 grams O2/32 grams = 11.1 moles O2 check for limiting reactant 11.1 moles O2 (2 mole H2/1 mole O2) = 22.2 mole H2 and H2 has no where near that many moles, so limits and drives reaction so, as they are one to one...... 22.2 moles of H2O are produced
210.3 moles of H2 are contained in one gallon of H2O
For every mole of H2 used in a reaction, one mole of H2O is produced, assuming the reaction is the complete combustion of hydrogen gas. This is because the balanced chemical equation for this reaction is: 2H2 + O2 -> 2H2O.
This reaction is:2 Al + 2 H2O + 2 NaOH = 2 NaAlO2 + 3 H2From 4 moles of Al 6 moles of hydrogen are obtained.
CO2 + H2 -> CO + H2O one to one here 30.6 moles H2O (1 mole H2/1 mole H2O) = 30.6 moles Hydrogen gas needed
CO2 + H2 <-> CO + H2O all one to one ( I assume that 99.1 is grams. Always units!!!! ) 99.1 grams H2O (1 mole H2O/18.016 grams)(1 mole H2/1 mole H2O) = 5.50 moles of hydrogen gas needed If that was 99.1 moles water vapor then it would take 99.1 moles hydrogen gas at a one to one ratio.
Here is the BALANCED reaction eq'n. 2H2(g) + O2(g) = 2H2O(l) The prefix number are the Molar Ratios 2:1::2 So oxygen is ONE(1) mole Hence hydrogen is TWO(2) moles It produces TWO(2) moles of water.
In the reaction between H2 and O2 to produce water, the molar ratio is 2:1. This means every 2 moles of H2 react with 1 mole of O2 to produce 2 moles of water. To find the amount of water produced, we need to calculate the number of moles of H2 (15.8 g H2 / molar mass of H2) and then use the molar ratio to determine the number of moles of water produced. Finally, convert moles of water to grams using the molar mass of water.
Oxygen limits the reaction, so......Balanced equation. 2H2 + O2 -> 2H2O 7.89 mole H2O (1 mole O2/2 mole H2O) = 3.95 mole oxygen gas needed ------------------------------------------