Q141CP - Chemistry: Molecular Nature Of Matter And Change

Question

When barium (Ba) reacts with sulfur (S) to form barium sulfide (BaS), each Ba atom reacts with an S atom. If 2.50 cm³ of Ba reacts with 1.75 cm³ of S, are there enough Ba atoms to react with the S atoms (d of Ba 3.51 g/cm³ ; d of S 2.07 g/cm³ )?

Step-by-Step Solution

Answer

There is not enough Ba for the reaction with sulfur atoms.

Step 1: Calculating the amount of S

There is a 1:1 ratio of the atom, it is concluded that the same amount of each substance is required for the reaction. 

Calculating the S amount,$$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{=}\mathbf{ }\mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{ }\mathbf{\times }\mathbf{Density}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{75}{\mathbf{cm}}^{\mathbf{3}}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{07}\frac{\mathbf{g}}{{\mathbf{cm}}^{\mathbf{3}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{62}\mathbf{g} \\ \mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{S}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{62}\mathbf{g}}{\mathbf{32}\mathbf{.}\mathbf{06}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{113}\mathbf{moles} \end{gathered}$$

Step 2: Calculating the amount of Ba

To find the amount of Ba,

$$\begin{gathered} \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba}\mathbf{=}\mathbf{Volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba}\mathbf{\times }\mathbf{Density}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}{\mathbf{cm}}^{\mathbf{3}}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{51}\frac{\mathbf{g}}{{\mathbf{cm}}^{\mathbf{3}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{775}\mathbf{g} \\ \mathbf{Moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ba}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Mass}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{\mathbf{8}\mathbf{.}\mathbf{775}\mathbf{g}}{\mathbf{137}\mathbf{.}\mathbf{33}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0664}\mathbf{moles} \end{gathered}$$

There is not enough Ba for the reaction with sulfur atoms.