Chemistry College

NEEDD HELP URGENTLY, NOBODY ELSE IS HELPING FFS
2.0 mol of Ca(OH)2 are mixed with 2.0 mol of HCl according to the following equation:
Ca(OH)2+2HCl=CaCl2+2H2O
a. Which chemical is in excess and which is limiting reactant?
b. What is the excess in grams?
c.Theoretically,how many moles of H20 will be produced?

Answers

Answer 1

Answer:

Explanation:

Limiting is HCl and excess is Ca(OH)2

excess is 296 grams Ca(OH)2

2 moles H2O will be formed

Related Questions

After addition of 20.00 mL of 0.500 M standard KOH solution to 10.00 mL of formic acid (HCOOH, Ka = 1.8 × 10-4), the equivalence point is reached. What is the molarity of the formic acid?

What is the pH at the equivalence point, based on the question above? Please make a suggestion for an appropriate indicator.

Answers

Answer: 3.79

Explanation: The balanced chemical equation for the reaction between formic acid (HCOOH) and KOH is:

HCOOH + KOH → HCOOK + H2O

We can use the stoichiometry of this reaction to calculate the number of moles of formic acid that reacted with the KOH:

moles of KOH = (20.00 mL)(0.500 mol/L) = 0.01000 moles

moles of HCOOH = moles of KOH

Therefore, the initial number of moles of formic acid is:

moles of HCOOH = (10.00 mL)(x mol/L) = 0.01000 moles

where x is the molarity of formic acid.

Solving for x, we get:

x = 1.00 M

Therefore, the molarity of the formic acid is 1.00 M.

At the equivalence point, all of the formic acid has reacted with the KOH, and the solution contains only the salt formed by the reaction, potassium formate (HCOOK). The pH at the equivalence point can be calculated using the equation for the salt hydrolysis constant:

Kb = Kw/Ka

where Kb is the base dissociation constant of the conjugate base (formate ion), Kw is the ion product constant for water (1.0 × 10^-14 at 25°C), and Ka is the acid dissociation constant of the acid (formic acid). Rearranging this equation, we get:

Kb/Ka = [OH^-][HCOO^-]/[HCOOH]

At the equivalence point, the concentration of the formate ion (HCOO^-) is equal to the concentration of the KOH added (0.01000 moles / 30.00 mL = 0.3333 M). We can assume that the concentration of the hydroxide ion (OH^-) is also equal to 0.3333 M, since KOH is a strong base and will dissociate completely. Substituting these values into the equation above, we get:

Kb/Ka = (0.3333)^2 / [HCOOH]

Solving for [HCOOH], we get:

[HCOOH] = (0.3333)^2 / (1.8 × 10^-4) = 6181.5 M

Taking the negative logarithm of this concentration, we get the pH at the equivalence point:

pH = -log[HCOOH] = -log(6181.5) = 3.79

Therefore, the pH at the equivalence point is 3.79.

Regenerate response

Please help almost due?

Answers

Answer:

-lithium

-atomic number

-mass number

-protons

Explanation:

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