The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
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
Sn=2
C and O=6
Im pretty confident yet I forgot a bit.
323.5g - 301.2g = 22.3g is the change in mass.
Explanation:
To solve this question, we need to use the following formula:
M = n/V
So:
M = ??
n = 2.634 mol
V = 25.2 L
M = 2.634/25.2
M = 0.105 mol/L
Answer: a. 0.105
