Chem question please help

Question

if 10.0g of hydrogen and 75.0g of oxygen are  together in a reaction tube, how much water produced? what other gas is found in the reaction tube(besides water vapor) after the reaction, and how much of this gas remains?

Answer 1

First, you must write the balanced equation.

2H₂ + O₂ ==> 2H₂O

Next, find which, if either reactant, is limiting.

moles H₂ present = 10.0 g x 1 mole/2 g = 5 moles H₂ present

moles O₂ present = 75.0 g x 1 mole/32 g = 2.34 moles O₂ present

From the mole ratio in the balanced equation, you need twice as much H₂ as you have O₂.  In this case, you do have that, so O₂ is limiting.  (An easy way to find limiting reactant is to divided moles of each by the coefficient in the balanced equation, and the one with the smallest value is limiting. Here that is 5 moles H₂/2 = 2.5 and 2.34 O₂/1 = 2.34.  O₂ is limiting).

 

How much water is produced?

This will be limited by the moles of O₂ present.  

2.34 moles O₂ x 2 H₂O/1 mole O₂ = 4.68 moles H₂O (x 18 g/mole = 84.2 g H₂O if you want the mass)

 

What other gas is found in the tube after the reaction?

Since H₂ gas is in excess, there will be some residual H₂ left over.

 

How much H₂ remains?

Started with 5 moles H₂ and used 2x2.34 moles = 4.68 moles (see ratio of 2 moles H₂ per 1 mole O₂)

Thus, 5 moles - 4.68 moles = 0.32 moles H₂ gas left over (0.64 g H₂ if you want the mass)