Write the reaction for the synthesis of water vapor from hydrogen gas and oxygen gas and balance it. If you used 0.022 mol hydrogen, how many mol of oxygen would you need in order for it to completely react and have no limiting reactant?

Can you please explain in detail how to solve this problem?

H2+O2 >> H20 balance...

2H2+O2 >>2H2O

so you use oxygen at half the molar rate of hydrogen....011moles O2

The synthesis reaction for water vapor from hydrogen gas (H2) and oxygen gas (O2) can be represented as follows:

2 H2 + O2 → 2 H2O

To balance the equation, we need to make sure that the number of atoms on both sides of the equation is the same. In this case, we have two hydrogen atoms on the left and four hydrogen atoms on the right side, so we need to balance the hydrogen atoms first by adding a coefficient of 2 in front of H2:

2 H2 + O2 → 2 H2O

Now, let's count the oxygen atoms. We have two oxygen atoms on the left side and two oxygen atoms on the right side, so the equation is now balanced.

To determine the number of moles of oxygen required to completely react with 0.022 mol of hydrogen, we can use the stoichiometry of the balanced equation. From the balanced equation, we can see that the molar ratio of O2 to H2 is 1:2.

Therefore, for every 2 moles of hydrogen, we need 1 mole of oxygen. Given that we have 0.022 mol of hydrogen, we can set up a proportion:

2 mol of hydrogen / 1 mol of oxygen = 0.022 mol of hydrogen / x mol of oxygen

We can solve for 'x' by cross-multiplying and simplifying:

2x = 0.022
x = 0.022 / 2
x = 0.011 mol of oxygen

Hence, you would need 0.011 mol of oxygen in order for it to completely react and have no limiting reactant.