The following shows the precipitation reaction of barium chloride (BaCl₂) and sodium hydroxide (NaOH): BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)
If you have to prepare the reactants by dissolving 1.00 g of BaCl₂ and 1.00 g of NaOH in water,
(a) What is the limiting reactant?
(b) How many grams of Ba(OH)₂ are produced?
(c) If your experiment produced 0.700 g of Ba(OH)₂, what is the percent yield of Ba(OH)₂?
(d) Based on this percent yield, how much limiting reactant should be used to achieve the target Ba(OH)₂ theoretical yield?

Answers

Answer 1
Answer:

Answer:

The answer to your question is:

a) BaCl2

b) 0.8208 g

c) yield = 85.3 %

d)

Explanation:

                     BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)

Data

a) 1 g of BaCl₂

   1 g of NaOH

MW BaCl2 = 137 + (35.5x2) = 208 g

MW NaOH = 23 + 16 + 1 = 40 g

                               208 g of BaCl2 -------------  1 mol

                                    1 g of BaCl2 -------------    x

                                  x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2

                                    40 g of NaOH ------------  1 mol

                                       1 g of NaOH ------------   x

                                 x = (1 x 1) / 40

                                 x = 0.025 mol of NaOH

The ratio BaCl2 to NaOH is 1:1 (in the equation)

But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.

b)

                    1 mol of BaCl2 -------------- 1 mol of Ba(OH)2

                    0.0048 mol     ---------------   x

                     x = (0.0048 x 1) / 1

                     x = 0.0048 mol of Ba(OH)2

MW Ba(OH)2 = 137 + 32 + 2 = 171 g

                     171 g of Ba(OH)2 -------------------- 1 mol

                      x                         --------------------  0.0048 mol

                     x = (0.0048 x 171) / 1

                     x = 0.8208 g

c)Data

Ba(OH)2 = 0.700 g

                   % yield = 0.700 / 0.8208 x 100

                   % yield = 85.3

d)

Sorry, i don't understand this question

       


Related Questions

The least condensed state of matter is
The usual units of density are : cm3/ gcm2/gg/cmNm
In chemical reactions, atoms area. created. c. rearranged. b. destroyed. d. neutralized.
A solution is made by dissolving 54.0 g of silver nitrate in enough water to make 350.0 ml. A 10.00 ml portion of this solution is then diluted to a final volume of 250.0 ml. What is the total concentration of ions present in the final solution?
How many moles of copper are equivalent to 3.44 × 1023 atoms of copper?

Which formula represents a molecule with the most polar bond?a. CO
b. NO
c.HI
d.HCI

Answers

D) HCl is the most polar bond. This is because the electronegativity difference between these 2 elements in this molecule is larger than the other 3 molecules given. The bigger the electronegativity difference, the more polar the bond.

Answer:

HCL

Explanation:

Reference Table S lists the electronegativity of each element. The greater the difference in electronegativity, the more polar the bond. The electronegativity of H is 2.2 and the electronegativity of Cl is 3.2; the difference is the greatest of the choices listed.

from castle learning

Make the following conversion.

60.2 mg = _____ g

Answers

Answer:

60.2 mg is equal to 0.0602 g.

Explanation:

Given = 60.2 mg

To find : 60.2 mg = ? g

Solution:

We know that in 1 gram there are 1000 milligrams.

1 g = 1000 mg

Then in , 1 mg = 0.001 g

So, 60.2 mg = 60.2 mg 0.001 g=0.0602 g

the question is 0.0602

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3(s)? 2NH3(g) + 2O2(g) NH4NO3(s) + H2O(l)Given:
NH4NO3(s): ∆Hf = -365.56 kJ ∆Sf = 151.08 J/K.
NH3(g): ∆Hf = -46.11 kJ ∆Sf = 192.45 J/K.
H2O(l): ∆Hf = -285.830 kJ ∆Sf = 69.91 J/K.
O2(g): ∆Hf = 0.00 kJ ∆Sf = 205 J/K.

Answers

∆G =  ∆H - T∆S

NH3:
∆G = -46.11x10^3 - (298)(192.45) = -103460.1 J
O2:
∆G = 0 - (298)(205) = -61090 J
NH4NO3:
∆G = -365.56x10^3 - (298)(151.08) = -410581.84 J
H2O:
∆G = -285.830x10^3 - (298)69.91) = -306663.18 J

∆Grex = ∆Gproducts - ∆Greactants
∆Grex = (-410581.84 +  -306663.18) - (-103460.1/2 + -61090/2)
∆Grex =-634969.97 J/mol = -634.97 kJ/mol

Which of the following reactions is endothermic?H2(g) + ½O2(g) → H2O(g), ΔH = -57.82 kcal
½N2(g) + O2(g) + 8.1 kcal → NO2(g)
½N2(g) + 3/2H2(g) → NH3(g) + 11.0 kcal
C(diamond) + O2(g) → CO2, ΔH = -94.50 kcal

Answers

Answer:

2) ½N2(g) + O2(g) + 8.1 kcal → NO2(g)

Explanation:

Chemical reactions in general can be classified as endothermic and exothermic reactions.

Endothermic reactions are accompanied by the absorption of energy. Here, the enthalpy (ΔH) change of the reaction is positive.

Exothermic reactions are accompanied by the release of energy. Here, the enthalpy (ΔH) change of the reaction is negative.

1)H2(g) + ½O2(g) → H2O(g), ΔH = -57.82 kcal

Here, ΔH is negative, hence exothermic

2)½N2(g) + O2(g) + 8.1 kcal → NO2(g)

Here, energy is being absorbed, hence endothermic

3) ½N2(g) + 3/2H2(g) → NH3(g) + 11.0 kcal

Here, energy is released, hence exothermic

4)C(diamond) + O2(g) → CO2, ΔH = -94.50 kcal

Here, ΔH is negative, hence exothermic

An exothermic reaction releases heat having a negative gross heat of reaction. The endothermic reaction, on the other hand, has a positive gross heat of reaction. Among the choices, the endothermic reaction is the third reaction having a positive heat of reaction.

How to calculate molar mass

Answers

 Molar masses of chemical compounds are equal to the sums of the molar masses of all the atoms in one molecule of that compound. After that, If you have a subscript in a chemical formula, then you multiply the number of atoms of anything next to that subscript by the number of the subscript. I'm kinda new to this stuff. Hopefully I helped :/
Molar mass is the sum of all atomic masses in a given compound in relative grams. To explain how to calculate the Molar mass, let me show you an example.

Calculate the Molar mass of H2O

now, atomic mass of H - 1
Atomic mass of O - 16

Now, molar mass = 1(2) + 16

= 2 +16
= 18 u = 18 grams (relative mass in grams)

Match the landform to its description.caldera
made of pieces of lava
volcanic soil
material fills in valleys
cinder cone
bowl-shaped depression
shield volcano
wide summit, gentle slope
lava plateau
rich in nutrients

Answers

Answer:

cinder cone: made of pieces of lava

shield volcano: wide summit, gentle slope

volcanic soil: rich in nutrients

caldera: bowl shaped depression

lava plateau: material fills in valleys

Explanation:

it’s right

Answer:

Caldera, Bowl-Shaped depression

Shield volcano, Wide summit, Gentle slope

Lava plateau, Material fills in valleys

Volcanic soil, Rich nutrients

Cinder Cone, Made of pieces of lava

Explanation:

Got it right on edge

Other Questions
100 Points and Brainliest to the correct answer, if you type a random answer I will reportHow can matter and energy be described and conserved in a variety of systems? Hypotheses: As you view each scenario, make a prediction about what will occur when prompted by the video. The Iced Tea Debate The Salty Soup Predictions—What do you think will happen? Materials: 1. Demonstrations: “The Iced Tea Debate” and “The Salty Soup.” Procedures: 1. Observe and analyze the following video demonstrations: “The Iced Tea Debate” and “The Salty Soup.” 2. Use the data table to record observations on physical and chemical change, and the conservation of matter and energy. Variables: List the variables for The Iced Tea Debate: Independent: Dependent: Control: List the variables for The Salty Soup: Independent: Dependent: Control: Data and Observations: Record your detailed observations and draw some brief conclusions in the table below. The Iced Tea Debate The Salty Soup Describe the physical changes you observed. Describe the chemical changes you observed. Describe the instances of conservation of matter and energy in each demonstration. Questions and Conclusion 1. How was matter and energy conserved in each demonstration? 2. What phase changes did you observe? 3. What kind/s of energy transfers did you notice in each scenario? 4. Identify an example of matter and energy conservation in the world around you. Conclusion: How did your observations support or contradict your predictions? Describe an experiment that could further explore physical or chemical change.