Answer:
0.0344 moles and 1.93g.
Explanation:
Molarity is defined as the ratio between moles of a solute (In this case, KOH), and the volume. With molarity and volume we can solve the moles of solute. With moles of solute we can find mass of the solute as follows:
<em>Moles KOH:</em>
15.2mL = 0.0152L * (2.26mol / L) = 0.0344moles
<em>Mass KOH:</em>
0.0344 moles * (56.11g/mol) = 1.93g of KOH
Format Method - Writing the symbol of the cation and then the anion. Add whatever subscripts in order to balance the charges.
Crisscross Method - The numerical value of the charge of each ion is crossed over and becomes the subscripts for the other ion.
Answer:
141g of CCl₄
Explanation:
First, we have to write the balanced equation.
CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)
We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.
V = 14.9 dm³ = 14.9 L
T = 21°C + 273.15 = 294.15 K
P = 1.48 atm
R = 0.08206 atm.L/mol.K
We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.
Answer:
The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3
Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3
Given: 0.600 moleCl2; 0.500 mole Al
Required: Excess reactant___; Number of moles of AlCl3 produced__
Solution: Use dimensional analysis using the mole-mole rel
0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al
0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2
Based on the given:
0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)
0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)
Answer: Excess reactant is Al; Limiting reactant is Cl2
The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3
It turns liquid to gas. Hoped this helped!!
