Question

3.0 g H₂ reacts with 30.0 g O₂ to form H₂O. What is the limiting reagent of reaction? Calculate the amount of H₂O formed. Also calculate the amount of excess reactant.

Answer 1

Following will be chemical reaction: 

For limiting reagent, 

4 g H, reacts with = 32 g O₂

3 g H2 will react with \( = \frac{32\times 3}{4} = 24 \ g \ O_2\)

But according to problem, we have 30 g 02. viz O₂ is in excess,

So, the limiting reagent for this reaction is ydrogen. 

4 g H₂ form = 36 g H₂O 

3 g H₂ will produce  = \(\frac{36\times 3}{4}\)

= 27 g H₂O