Answer:
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
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
Answer:
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
