CHEMISTRY HIGH SCHOOL

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3 ( s ) + 2 HCl ( aq ) ⟶ CaCl2 ( aq ) + H 2 O ( l ) + CO 2 ( g ) How many grams of calcium chloride will be produced when 30.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid

Answers

Answer 1

Answer:

Answer: The mass of $CaCl_2$ produced is, 15.2 grams.

Explanation : Given,

Mass of $CaCO_3$ = 30.0 g

Mass of $HCl$ = 10.0 g

Molar mass of $CaCO_3$ = 100 g/mol

Molar mass of $HCl$ = 36.5 g/mol

First we have to calculate the moles of $CaCO_3$ and $HCl$ .

$\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}$

$\text{Moles of }CaCO_3=\frac{\text{Given mass }CaCO_3}{\text{Molar mass }CaCO_3}=(30.0g)/(100g/mol)=0.300mol$

and,

$\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}$

$\text{Moles of }HCl=\frac{\text{Given mass }HCl}{\text{Molar mass }HCl}=(10.0g)/(36.5g/mol)=0.274mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $HCl$ react with 1 mole of $CaCO_3$

So, 0.274 moles of $HCl$ react with $(0.274)/(2)=0.137$ moles of $CaCO_3$

From this we conclude that, $CaCO_3$ is an excess reagent because the given moles are greater than the required moles and $HCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CaCl_2$

From the reaction, we conclude that

As, 2 mole of $HCl$ react to give 1 mole of $CaCl_2$

So, 0.274 mole of $HCl$ react to give $(0.274)/(2)=0.137$ mole of $CaCl_2$

Now we have to calculate the mass of $CaCl_2$

$\text{ Mass of }CaCl_2=\text{ Moles of }CaCl_2* \text{ Molar mass of }CaCl_2$

Molar mass of $CaCl_2$ = 110.98 g/mole

$\text{ Mass of }CaCl_2=(0.137moles)* (110.98g/mole)=15.2g$

Therefore, the mass of $CaCl_2$ produced is, 15.2 grams.

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Regarding water's ability to move up a glass tube, which of the following is true?

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A vineyard has 145 acres of Chardonnay grapes. A particular soil supplement requires 5.50 g for every square meter of vineyard. How many kilograms of the soil supplement are required for the entire vineyard? (Recall that 1 km2 = 247 acres.) Express your answer in kilograms to three significant figures.

Answers

1.36.10 ^ 6 kg of supplements are required for the entire vineyard

Further explanation

7 main quantities have been determined based on international standards, namely:

1. Length, meters (m)

2. Time, second (s)

3. Mass, kilograms (kg)

4. Temperature, kelvin (K)

5. Light intensity, candela (cd)

6. Electric current, ampere (A)

7. Amount of substance, mol (m)

Derivative magnitude is a quantity derived from one or more principal quantities. So in addition to 7 principal quantities, other quantities are derived quantities

An area with the formula length x width is a unit derived from the length of the principal. The international standard unit is square meters (m²).

other area units: Km², hm², dam², m², dm², cm², and mm²

Hectare is an SI unit

1 Hectare is equal to 100 a (Are) or 10000 m² (ten thousand square meters) or 100 x 100 m 1 Hectares = 2.47 acres

In the question , there is a 145 acres vineyard, with the supplement is given 5.5 gram / m² vineyard.

So for 1 km², a supplement =

1 km² = 10⁶ m²

5.5 gram / m² = 5.5.10⁶ grams / km²

whereas 1 km² = 247 acres and vineyard = 145 acres so

$5.5.10^6*(145)/(247)$

= 3.2287.10⁶ grams

= 3.23.10³ kg (3 significant numbers: 3,2 and 3)

Learn more

Convert the following metric units of weight

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conversion factor

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convert a mass of 2.93 pounds to ounces

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Keywords: area, convert, acres

The soil supplement required for the entire vineyard is $\boxed{3.23 * {{10}^3}{\text{ kg}}}$ .

Further Explanation:

There are two types of units. One is basic or fundamental while the other ones are derived units. Basic units cannot be further reduced and other quantities are expressed in these units. Derived units are those that can need basic units to express themselves. Area, density, volume and velocity are some examples of derived units.

Seven basic units are present in the SI system. These are as follows:

1. Meter (m)

2. Kilogram (kg)

3. Second (s)

4. Kelvin (K)

5. Ampere (A)

6. Mole (mol)

7. Candela (Cd)

Firstly, the area of vineyard has to be converted into ${\text{K}}{{\text{m}}^2}$ . The conversion factor for this is,

$1{\text{ acre}} = \left( {\frac{1}{{247}}} \right){\text{ K}}{{\text{m}}^2}$

Therefore the area of vineyard can be calculated as follows:

$\begin{aligned}{\text{Area of vineyard}} &= \left( {145{\text{ acres}}} \right)\left( {\frac{{1/247{\text{ K}}{{\text{m}}^2}}}{{1{\text{ acre}}}}} \right)\n&= 0.587045{\text{ K}}{{\text{m}}^2}\n\end{aligned}$

The area is again converted into ${{\text{m}}^{\text{2}}}$ . The conversion factor for this is,

$1{\text{ K}}{{\text{m}}^2} = {10^6}{\text{ }}{{\text{m}}^2}$

So the area of vineyard can be calculated as follows:

$\begin{aligned}{\text{Area of vineyard}}&= \left( {0.587045{\text{ K}}{{\text{m}}^2}}\right)\left( {\frac{{{{10}^6}{\text{ }}{{\text{m}}^2}}}{{1{\text{ K}}{{\text{m}}^2}}}}\right)\n&= 587045{\text{ }}{{\text{m}}^2}\n\end{aligned}$

The amount of supplement required for the entire vineyard can be calculated as follows:

$\begin{aligned}{\text{Amount of supplement required}}&= \left( {587045{\text{ }}{{\text{m}}^2}} \right)\left( {\frac{{5.50{\text{ g}}}}{{1{\text{ }}{{\text{m}}^2}}}} \right)\n&=3.2287475 * {10^6}{\text{ g}}\n\end{aligned}$

The amount of supplement is to be converted into kg. The conversion factor for this is,

$1{\text{ g}} = {\text{1}}{{\text{0}}^( - 3)}{\text{ kg}}$

Therefore the amount of supplement can be calculated as follows:

$\begin{aligned}{\text{Amount of supplement}}&= \left({3.2287475 * {{10}^6}{\text{ g}}} \right)\left( {\frac{{{{10}^( - 3)}{\text{ kg}}}}{{1{\text{ g}}}}} \right)\n&= 3.2287475 * {10^3}{\text{ kg}}\n&\approx 3.23 * {10^3}{\text{ kg}}\n\end{aligned}$

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Basic concepts of chemistry

Keywords: supplement, vineyard, 5.50 g, 3.23*10^3 kg, conversion factor, basic units, fundamental units, derived units, area, volume, density, kg, m, Km, acre.

Infrequent participation in physical activity can lead toa. high blood pressure during exercise
b. muscle strain and soreness after exercise
c. extra calories burned during exercise
d. low heart rate after exercise

Answers

We can see here that infrequent participation in physical activity can lead to: b. muscle strain and soreness after exercise

What is physical activity?

Infrequent participation in physical activity can lead to muscle strain and soreness after exercise. When individuals are not regularly engaged in physical activity, their muscles may not be conditioned or accustomed to the stress and demands placed on them during exercise. As a result, they may experience muscle strain, micro-tears in the muscle fibers, and delayed-onset muscle soreness (DOMS) following physical activity.

It's worth mentioning that leading a sedentary lifestyle, which involves minimal physical activity overall, can have broader health implications beyond muscle strain and soreness.

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You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain 69 quibs of a substance called skvarnick. You can trade this Skvarnick for gold coins, but the vendors all measure skvarnick units of sleps; not quibs. 9 sleps is equal to 6 quibs. If you have 14 quibs of skvarnick, how many sleps do you have? Round your answer to the nearest tenth (one decimal place). Type only the number; not the number and unit.

Answers

Answer: 21 sleps

Explanation:

Given information: The substance skvarnick can be traded for gold coins in units of sleps and not quibs. Thus we need to convert the units of quibs to sleps.

It is given that 9 sleps is eqaul to 6 quibs, 14 quibs = ____ sleps.

$6\text{ quibs} = 9\text{ sleps}$

$14\text{ quibs} = (9)/(6)* {14}\text{ sleps}$

$=21\text{ sleps}$

The cell theory teaches that _____.the basic unit of life is the organism
the basic unit of life is the cell
all cells have the same basic structure
cells are miniature reflections of the organisms from which they are taken

Answers

The cell theory teaches that :

the basic unit of life is the cell
Explanation:

In biology, a scientific theory is that the historical theory, currently universally accepted, that living organisms square measure created from cells, that they're the fundamental structural/organizational unit of all organisms, which all cells come back from pre-existing cells. Cells square measure the fundamental unit of structure all told organisms and conjointly the fundamental unit of the replica.

Answer:

the basic unit of life is the cell

I have to write a 150 word paragraph or two that describes at least three everyday things that exist or occur because of science. I also have to make sure i use examples. I don't have a clue where to start ??? PLEASE help!

Answers

You can do something like natural disasters and explain why they happen

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