Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2
Balance the reaction so the same number of each type of atom is on each side. Which option seems balanced to you?
Molarity is moles of solute per litre of solvent. We're given mass of solute, but can convert to moles using the molar mass of NaOH (40 g/mol). There are 10 g / 40 g/mol = 0.25 mol of NaOH in the solution.
M = 0.25 mol / 2.0 L = 0.125M
The second answer (0.13M) is the correct one.
Answer:
You need 375 mL of BaCl2 solution.
Explanation:
M1V1=M2V2
Dilution formula. Substitute known values and solve for V1.
M1 = 2.0 M
M2 = 1.50 M
V2 = 500 mL
(2.0 M)(V1) = (1.50 M)(500 mL)
V1 = (1.50 M)(500 mL) / (2.0 M)
V1 = 375 mL
To get radius, divide the diameter into 2. 3.6÷2 = 1.8 cm
