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

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?

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