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2 years ago

What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?

1 answer:

8 0

When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) * (1L ca(OH)2/0.585 mol Ca(OH)2

= 20.4 L

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According to newtons third law forces always occur in equal but____ pairs?

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2 years ago

199.5 grams unrefined dark crystalline sugar to cups

Dmitry [639]

Answer:

0.9975 cup

Step-by-step explanation:

"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."

200.0 g brown sugar = 1 cup

199.5 g brown sugar = 199.5× 1/200 .0

199.5 g brown sugar = 0.9975 cup

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2 years ago

The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of

Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be

10 g

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of

1 g

of that substance by

∘

Water has a specific heat of approximately

4.18

∘

. This tells you that in order to increase the temperature of

1 g

of water by

∘

, you need to provide

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of

1 g

of water by

∘

Well, you'd need

4.18 J

to increase it by

∘

, another

4.18 J

to increase it by another

∘

, and so on. This means that you'd need

4.18 J

41.8 J

to increase the temperature of

1 g

of water by

∘

Now look at the value given to you. If you need

41.8 J

to increase the temperature of

1 g

of water by

∘

, what mass of water would require

times as much heat to increase its temperature by

∘

1 g

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

⋅

, where

- heat absorbed/lost

- the mass of the sample

- the specific heat of the substance

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

⋅

4.18

∘

⋅

(

−

)

∘

418

4.18

⋅

10 g

5 0

2 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

2 years ago

Read 2 more answers