Answer:
240.17 g Ba3(PO4)2
Explanation:
1. Determine the limiting reactant.
2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
moles H3PO4 = M x V = 3 x 0.286 = .858 moles H3PO4
moles Ba(OH)2 = M x V = 1.4 x 0.855 = 1.197 moles Ba(OH)2
ratio Ba(OH)2 : H3PO4 = 1.197: .858 = 1.39: 1
stoichiometric ratio Ba(OH)2 : H3PO4 = 3:2
Ba(OH)2is the limiting reactant
MM Ba3(PO4)2 = 601.92 g/mol
g Ba3(PO4)2 = moles Ba(OH)2 x(1 mol Ba3(PO4)2/3 moles Ba(OH)2) x (MM Ba3(PO4)2/ 1mol Ba3(PO4)2) = 1.197 x 1/3 x 601.92 = 240.17 g Ba3(PO4)2
In general oxidation is defined as gain of oxygen or loss of electron or hydrogen by an atom. Reduction is defined as gain of electron or hydrogen or loss of oxygen by an atom.
In a balanced redox reaction we have two half reactions
a) reduction half reaction : the oxidation number of element decreases
b) oxidation half reaction : the oxidation number of element increases
The element undergoing reduction is Pd.
the oxidation number of Pd decreases from +2 to 0
Thus the reduction half reaction will be
The Pd (II) ion will take two electrons and will give Pd (0)
Answer is
Volume = a x a x a
V = 2 cm x 3 cm x 4 cm => 24 cm³
Density = 19.3 g/cm³
Mass = ?
Therefore:
m = D x V
m = 19.3 x 24
m = 463.2 g
The formula of density is mass / volume
This means that
- high mass, low volume = high density
- high mass, high volume = so-so
- low mass, high volume = low density
From the graph shown,
D has the lowest density because it has low mass yet high volume.
