Question

Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).

Answer 1

Answer: 7.29 g of $H_2S$ will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

${\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles$

$\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles$

$8H_2+S_8\rightarrow 8H_2S$

According to stoichiometry :

1 mole of $S_8$ require = 8 moles of $H_2$

Thus 0.0267 moles of $S_8$ will require=$\frac{8}{1}\times 0.0267=0.214moles$  of $H_2$

Thus $S_8$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

As 1 mole of $S_8$ give = 8 moles of $H_2S$

Thus 0.0267 moles of $S_8$ give =$\frac{8}{1}\times 0.0267=0.214moles$  of $H_2S$

Mass of $H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g$

Thus 7.29 g of $H_2S$ will be produced from the given masses of both reactants.