Answer:
113 g NaCl
Explanation:
The Ideal Gas Law equation is:
PV = nRT
In this equation,
> P = pressure (atm)
> V = volume (L)
> n = number of moles
> R = 8.314 (constant)
> T = temperature (K)
The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.
(1.50 atm)(15.0 L) = n(8.314)(280. K)
2250 = n(2327.92)
0.967 moles F₂ = n
Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
1 F₂ + 2 NaCl ---> Cl₂ + 2NaF
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl
Answer:
101 L
Explanation:
35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O
35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O
35.0 g KOH is the limiting reactant
Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of 
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get
To calculate the volume of acid, we use the equation given by neutralisation reaction:
where,
are the molarity and volume of stock acid which is
are the molarity and volume of dilute acid which is
We are given:
Putting values in above equation, we get:
Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
True will end up being the answer
