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.

Step-by-step explanation:

To calculate the moles :

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

`{\text{Moles of} H_2}=(1.35g)/(2.01g/mol)=0.672moles`

`\text{Moles of} S_8=(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=
`(8)/(1)* 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 =
`(8)/(1)* 0.0267=0.214moles` of
`H_2S`

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

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