Calculate the amount of H2 which is left unreacted in the given reaction 2H2 + O2 → 2H2O If 8 g of H2 is mixed with 16 g O2?

Question

Calculate the amount of H2 which is left unreacted in the given reaction 2H2 + O2 → 2H2O If 8 g of H2 is mixed with 16 g O2? (A) 3g (B) 6g (C) 1g (D) 4g

Tardigrade Admin · Accepted Answer

beginmatrixGiven moles&&((8/2))&&((16/32)) &&4&&0.5 endmatrix So O2 is the limiting reagent moles used of H2 = 1 So unreacted H2 = 6 gm

Q. Calculate the amount of $H_{2}$ which is left unreacted in the given reaction
$2 H_{2} + O_{2} \to 2 H_{2} O$
If $8 g$ of $H_{2}$ is mixed with $16 g O_{2}$ ?

$3 g$

$6 g$

1g

4g

$G i v e n m o l es (\frac{8}{2}) 4 (\frac{16}{32}) 0.5$
So $O_{2}$ is the limiting reagent
moles used of $H_{2} = 1$
So unreacted $H_{2} = 6 g m$

Q. Calculate the amount of H2​ which is left unreacted in the given reaction 2H2​+O2​→2H2​O If 8g of H2​ is mixed with 16gO2​?

3g

6g

1g

4g

Givenmoles​​(28​)4​​(3216​)0.5​ So O2​ is the limiting reagent moles used of H2​=1 So unreacted H2​=6gm

Q. Calculate the amount of $H_{2}$ which is left unreacted in the given reaction
$2 H_{2} + O_{2} \to 2 H_{2} O$
If $8 g$ of $H_{2}$ is mixed with $16 g O_{2}$ ?

$3 g$

$6 g$

$G i v e n m o l es (\frac{8}{2}) 4 (\frac{16}{32}) 0.5$
So $O_{2}$ is the limiting reagent
moles used of $H_{2} = 1$
So unreacted $H_{2} = 6 g m$