Question

One of relatively few reactions that takes place directly between two solids at room temperature is
`Ba(OH)_2.8H_2O(s)+NH_4SCN(s) \longrightarrow Ba(SCN)_2(s)+H_2O(l)+NH_3(g)`In this equation, the .8H₂O in Ba(OH)₂.8H₂O indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balanced the equation. b. What mass of ammonium thiocyanate (NH₄SCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

Answer 1

Step-by-step explanation:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

Making the number of atom on the RHS = number of atoms on the LHS

Balanced equation:

Ba(OH)₂.8H₂O(s) + 2NH₄SCN(s) → Ba(SCN)₂(s) + 10H₂O(l) + 2NH₃(g)

B.

Given:

Mass of Ba(OH)₂.8H₂O = 6.5 g

Molar mass = 137 + (16 + 1) × 2 + 8 × (2 + 16)

= 315 g/mol

Number of moles = mass/molar mass

= 6.5/315

= 0.0206 moles

By stoichiometry, since 1 mole of Ba(OH)₂.8H₂O reacted with 2 mole of NH₄SCN. Therefore, number of moles of NH₄SCN = 2 × 0.0206

= 0.0413 moles

Molar mass of NH₄SCN = (14 × 2) + (4 × 1) + 12 + 32

= 76 g/mol

Mass = 76 × 0.0413

= 3.137 g of NH₄SCN

Answer 2

a) Ba(OH)₂.8H₂O(s) + 2 NH₄SCN(s) → Ba(SCN)₂(s) +10 H₂O(l) + 2 NH₃(g)

b) 3.14g must be added

Step-by-step explanation:

a) For the reaction:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +10 H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + 2 NH₄SCN(s) → Ba(SCN)₂(s) +10 H₂O(l) + 2 NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) = 0.0206moles of Ba(OH)₂.8H₂O. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = 0.0412moles of NH₄SCN. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = 3.14g must be added