Question

2H2S (g) + 3O2 (g) → 2H2O (g) + 2SO2 (g)

If 3.50 g of H2S are used in the above reaction, what will be the theoretical yield of water in grams?

Answer 1

there are three steps to do: 1) convert grams to moles, 2) convert moles to moles and 3) convert moles to grams

1) let's convert the 3.50 grams of H₂S to moles using the molar mass

molar mass H₂S= (2 x 1.01) + 32.0= 34.02 g/mol

3.50 g (1 mol/ 34.02 g)= 0.103 mol H₂S

2) let's convert moles of H₂S to moles of water using the mole-mole ratio from the balanced equation.

0.103 mol H₂S (2 mol H₂O/ 2 mol H₂S)= 0.103 mol H₂O

3) lastly, let's convert moles of water to grams using the molar mass

molar mass of H₂O= (2 x 1.01) + 16.0= 18.02 g/mol

0.103 mol H₂O (18.02 g/ mol)= 1.85 grams of H₂O

Answer 2

The theoretical or the reaction yield is the amount of the moles formed by the consumption of the reactants. The theoretical yield of the water is 1.85 gm.

What are moles and masses?

Moles and the mass of the solute or substance is the quantity of the atoms and the molecules present or the weight of the compound.

Calculate the moles of the hydrogen sulfide as:

$\begin{aligned}\rm moles &= \rm \dfrac{mass}{\rm molar \; mass}\\\\&= \dfrac{3.50}{34.02}\\\\&= 0.103 \;\rm mol\end{aligned}$

Now calculate the moles of the water from hydrogen sulfide as,

Initially 2 moles hydrogen sulfide = 2 moles of water

So, 0.103 moles of hydrogen sulfide = 0.103 moles of water

Finally convert the moles of water to the mass as:

$\begin{aligned} \rm mass &= \rm moles \times molar\; mass\\\\&= 0.103 \times 18.02\\\\&= 1.85\;\rm gm\end{aligned}$

Therefore, 1.85 gm of water is produced.

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