Question

One of relatively few reactions that takes place directly between two solids at room temperature is Ba(OH)2 · 8H2O + ammonium thiocyanate --> Barium thiocyanate + water + ammonia In this equation, the · 8H2O in Ba(OH)2 · 8H2O indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. What mass of ammonium thiocyanate must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

Answer 1

Answer:

$\boxed{\text{3.1 g}}$

Explanation:

We will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:           315.46               76.12

           Ba(OH)₂·8H₂O + 2NH₄SCN ⟶ Ba(SCN)₂ + 2NH₃ + 10H₂O

m/g:             6.5

1. Moles of Ba(OH)₂·8H₂O

$\text{Moles of Ba(OH) _{2}\cdot 8H _{2} O}\\= \text{ 6.5 g Ba(OH) _{2}\cdot 8H _{2} O} \times \dfrac{\text{1 mol Ba(OH) _{2}\cdot 8H _{2} O}}{\text{ 315.46 g Ba(OH) _{2}\cdot 8H _{2} O}}\\= \text{0.0206 mol Ba(OH) _{2}\cdot 8H _{2} O}$

2. Moles of NH₄SCN

The molar ratio is 2 mol NH₄SCN:1 mol Ba(OH)₂·8H₂O

$\text{Moles of NH _{4} SCN} =\text{0.0206 mol Ba(OH) _{2}\cdot 8H _{2} O} \times \dfrac{\text{2 mol NH _{4} SCN}}{\text{1 mol Ba(OH) _{2}\cdot 8H _{2} O}}\\= \text{0.0412 mol NH _{4} SCN}$

3. Mass of NH₄SCN

$\text{Mass of NH _{4} SCN} = \text{0.0412 mol NH _{4} SCN} \times \dfrac{\text{76.12 g NH _{4} SCN}}{\text{1 mol NH _{4} SCN}} =\\\textbf{3.1 g NH _{4} SCN}\\\\\text{You must use }\boxed{\textbf{3.1 g}}\text{ NH _{4} SCN}$