Answer:
mNaNO3 =765g
Explanation:
First, we write the balanced chemical equation representing the chemical reaction that happened between aluminum nitrate and sodium chloride.
Balanced chemical equation:
AL(NO3)3+3NaCl ⟶ 3 NaNO3+AlCl3
According to the equation, each mole of aluminum nitrate requires three moles of sodium chloride. Thus, the required number of moles of sodium chloride is
4 Mol ⋅ 3 = 12mol
Based on the data provided in the table, there were 9 moles of sodium chloride used in the reaction, which was not enough for the entirety of aluminum nitrate to react. So, sodium chloride must have been the limiting reactant.
Therefore, we use the number of moles (n) of sodium chloride to calculate the number of moles of sodium nitrate, which has a 1:1 ratio with sodium chloride.
Number of moles sodium nitrate:
nNaNO3=nNaCl
nNaNO3 = 9 mol
We can also calculate the mass (m) of sodium nitrate that was produced by multiplying its number of moles by its molar mass (MM), 85.00g/mol.
Mass of sodium nitrate produced:
mNaNO3 = nNaNO3 ⋅ MMNaNO3
mNaNO3 = 9 mol ⋅ 85.00 g/mol
mNaNO3 =765g
Answer:
SN1 = a stepwise loss of a leaving group to form a carbocation followed by nucleophilic attack
Explanation:
Since 2-methyl-2-butanol is tertiary alcohol, the first step will be loss of leaving group to form a 3° carbocation which is very stable and favours SN1, followed by attack of nucleophile
Answer:
1. 0.45 mole
2. 49.95g
Explanation:
The following were obtained from the question:
Volume of solution = 300mL = 300/1000 = 0.3L
Molarity = 1.5 M
Mole of CaCl2 =?
1. We can obtain the mole of the solute as follow:
Molarity = mole of solute /Volume of solution
1.5 = mole of solute/0.3
Mole of solute = 1.5 x 0.3
Mole of solute = 0.45 mole
2. The grams in 0.45 mole of CaCl2 can be obtained as follow:
Molar Mass of CaCl2 = 40 + (35.5 x 2) = 40 + 71 = 111g/mol
Mole of CaCl2 = 0.45 mole
Mass of CaCl2 =?
Mass = number of mole x molar Mass
Mass of CaCl2 = 0.45 x 111
Mass of CaCl2 = 49.95g
