Reactants:

hydrogen sulfide → H_{2}S

sodium hydroxide → NaOH

Products:

sodium sulfide → Na_{2}S

water → H_{2}O

**Balanced reaction: H _{2}S + 2 NaOH → Na_{2}S + 2 H_{2}O**

**Calculate theoretical yield:**

**Na _{2}S**

molar mass Na_{2}S = 78.04 g/mol

molar mass H_{2}S = 34.08 g/mol

$\mathbf{mass}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}\mathbf{S}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}}}{\mathbf{34}\mathbf{.}\mathbf{08}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}\mathbf{S}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}}}\mathbf{\times}\frac{\mathbf{78}\mathbf{.}\mathbf{04}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}\mathbf{S}}{\overline{)\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}\mathbf{S}}}$

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water.

How many grams of sodium sulfide are formed if 1.60 g of hydrogen sulfide is bubbled into a solution containing 2.13 g of sodium hydroxide, assuming that the sodium sulfide is made in 93.0 % yield?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Percent Yield concept. You can view video lessons to learn Percent Yield. Or if you need more Percent Yield practice, you can also practice Percent Yield practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Zellmer's class at OSU.