1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
eimsori [14]
2 years ago
5

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Chemistry
1 answer:
Brrunno [24]2 years ago
6 0

Answer:

Approximately 81.84\%.

Explanation:

Balanced equation for this reaction:

{\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm  NaCl}\, (aq) + {\rm CaCO_{3}}\, (s).

Look up the relative atomic mass of elements in the limiting reactant, \rm CaCl_{2}, as well as those in the product of interest, \rm CaCO_{3}:

  • \rm Ca: 40.078.
  • \rm Cl: 35.45.
  • \rm C: 12.011.
  • \rm O: 15.999.

Calculate the formula mass for both the limiting reactant and the product of interest:

\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}.

\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the quantity of the limiting reactant (\rm CaCl_{2}) available to this reaction:

\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}.

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant (\rm CaCl_{2}) and the product ({\rm CaCO_{3}}) are both 1. Thus:

\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1.

In other words, for every 1\; \rm mol of \rm CaCl_{2} formula units that are consumed, 1\; \rm mol\! of \rm CaCO_{3} formula units would (in theory) be produced. Thus, calculate the theoretical yield of \rm CaCO_{3}\! in this experiment:

\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}.

Calculate the theoretical yield of this experiment in terms of the mass of \rm CaCO_{3} expected to be produced:

\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}.

Given that the actual yield in this question (in terms of the mass of \rm CaCO_{3}) is 13.19\; \rm g, calculate the percentage yield of this experiment:

\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}.

You might be interested in
Help me please!! :D​
dem82 [27]

Answer: Solar eclipse

Explanation:

4 0
3 years ago
A gas at 29.4 kPa is cooled from a temperature of 75°C to a temperature of 25°C at constant volume. What is the new pressure of
kirill115 [55]

<span>To solve this we assume that the gas inside the balloon is an ideal </span>gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant volume pressure and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

 

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 25 x 29.4 / 75

P2 = 9.8 kPa

7 0
3 years ago
Read 2 more answers
If the temperature of a piece of steel decreases, what happens to its density? a. The density decreases. b. The density increase
vampirchik [111]

Answer:

Explanation:

4 0
3 years ago
Read 2 more answers
Choose the molecular compound among the substances listed below.
agasfer [191]

Answer: Among the listed substances XeCl_{4} is the molecular compound.

Explanation:

A chemical compound formed by the chemical combination of two or more non-metals is called a molecular compound or covalent compound.

For example, Xe and Cl are non-metals. The compound formed by them is XeCl_{4} which is a molecular compound.

A molecular compound is formed by sharing of atoms between the combining atoms.

Whereas NaF, MnCl_{2} and CaO are all ionic compounds as they are formed by chemical combination of a metal and a non-metal.

Thus, we can conclude that among the listed substances XeCl_{4} is the molecular compound.

4 0
2 years ago
Although I am solid, I am so light that I can float in any liquid listed on the chart. What am I?-----
evablogger [386]

Answer:

I know that Aerogel is the lightest metal in existence, but I don't think it would help much with your answer. I mean you can give it a try?

3 0
3 years ago
Other questions:
  • plate a and plate b are both continental plates. what is ghe most likely result if the two plates collide?
    8·1 answer
  • What is the conecntration of Fe3 and the concentration of No3- present in the solution that result when 30.0 ml of 1.75M Fe(No3)
    10·1 answer
  • Did dalton tried to describe the atom structure?​
    15·1 answer
  • 1. What controls the cell?<br><br> 2. Calls can be many ____ and _____
    7·1 answer
  • Explain, in terms of ions, why the ability to conduct an electric current is greater for the
    11·1 answer
  • Define natural resources
    9·1 answer
  • Imagine cutting a bar of pure gold in half, and then cutting it in half again. Could you divide the gold like this forever?
    11·1 answer
  • PLEASE HELP ASAP
    7·1 answer
  • For a wind farm to be effective, it needs to be in a place with what kind of wind?
    7·2 answers
  • Density and Years: What do you know?Objectives: determine and compare density of pennies to assess composition. Procedure:1. Rea
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!