Answer:
A double displacement is a type of chemical reaction that in which the reactant ions exchange places to form new products
Explanation:
I took the quiz
Answer:
When two different atoms or groups of atoms are exchanged, then it is a double displacement reaction.
(3) energy, charge and mass
(4) energy, charge and volume
Answer:
Explanation:
A pure substance is a substance that consists of only one type of particle, either atoms, molecules, or ions, and has a consistent and uniform composition throughout. Pure substances cannot be separated into other substances by physical means. They have well-defined chemical properties and distinct physical properties, such as melting point and boiling point.
Two examples of pure substances are:
Elemental Hydrogen (H2): Elemental hydrogen consists of diatomic molecules, each containing two hydrogen atoms bonded together. It is a pure substance because it consists only of hydrogen molecules and cannot be separated into other substances without chemical reactions.
Water (H2O): Water is a pure substance composed of water molecules, each consisting of two hydrogen atoms and one oxygen atom bonded together. It is also a pure substance because it has a consistent and uniform composition throughout and cannot be separated into other substances without breaking its molecular bonds.
These examples illustrate the concept of pure substances, where the composition is homogeneous and consistent throughout the entire sample.
A. What is the limiting reagent?
B. How many grams of the excess reactant remains?
C. How many grams of each product is formed?
D. If 12 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
Answer:
Explanation:
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
MM for each compound -
Pb(NO3): 207 + 14x2 + 16x3x2 = 331
PI2: 207 + 127x2 = 461
NaI: 23 + 127 = 150
NaNO3: 23 + 14 + 16x3 = 85
Moles of Pb(NO3)2 = 50/331 = 0.15
Moles of NaI = 30/150 = 0.2
Ratio of moles is 1:2
So NaI is limiting
Limited to 0.2/2 = 0.1 mole of Pb(NO3)2
Excess = 0.15 - 0.1 = 0.05 mole
Mass remains = 0.05x331 = 16.55 grams
Moles of NaNO3 formed = Moles of NaI reacted = 0.2
Mass = 0.2x85 = 17 grams
Moles of PbI2 formed = Moles of Pb(NO3)2 reacted = 0.1
Mass = 0.1x461 = 46.1 grams
If 12 grams of NaNO3 actually formed in the reaction,
percent yield = 12/17x100% = 70.6%
Answer:
Explanation:
Moles of Pb(NO3)2 = mass/molecular mass
= 50.0 grams/(207.20*1 + 14.01*2 + 16*6)
= 50.0 grams/331.22
= 0.15 moles
Moles of NaI
= 30/(22.99+126.9)
= 30/149.89
= 0.2 Moles
A. NaI is less 2x Pb(NO3)2 so NaI is the limiting reagent.
B. The ratio is 1 to 2 so there is 0.15 - 0.2/2 = 0.05 mole
or 16.78 grams of Pb(NO3)2 left.
C. As NaI is limiting, only 0.2 Moles of NaNO3 is formed.
Mass = Moles * Molecular Mass
Molecular Mass of NaNO3 can be calculated as:
Na - 22.99
N - 14.01
O - 3(16) = 48
23+14+48 = 85gram / mole
Thus, Mass = 0.2*85 = 17 gram of NaNO3
Mass is conserved in a chemical reaction.
Mass of PbI2 can be calculated as:
50+30-16.78-17
= 46.3 gram of PbI2
Mass =
12.75
Thus, 12.75g of Sodium Nitrate can be formed