Question

3 g of H₂ is mixed with 29 g of O₂ to yield water. 

1. Which is the limiting reagent? 

2. Calculate the maximum amount of water that can be formed. 

3. Calculate the amount of the reactants which remains unreacted.

Answer 1

1.

According to the equation, 4g H₂ requires 32g 

So 3g H₂ requires \(\frac{377\times33}{44}\)= 24g O₂ . 

Here 3g H₂ is mixed with 29g of O₂ . 

All H₂ will react. Hence H₂ is the limiting reagent. 

2. According to the equation, 4g H gives 36g H₂O. Hence 3g H₂ will give 36 × \(\frac{3}{4}\) = 27g H₂O.

3. Amount of O₂ unreacted = (29 – 24)g = 5g.