All categories

3 years ago

Find the mass of calcium oxide produced from the thermal decomposition of 50 g of calcium carbonate.

Chemistry

1 answer:

anastassius [24]3 years ago

3 0

Answer:

28 g

Explanation:

The chemical equation of the reaction is given below:

CaCO₃ (s) ⇒ CaO (s) + O₂ (g)

From the chemical equation, 1 mole of CaCO₃ produces 1 mole of CaO.

So, we find the number of moles,n of CaCO₃ in 50 g from

n = m/M where n = number of moles of CaCO₃, m = mass of CaCO₃ and M = molar mass of CaCO₃ = 40 g/mol + 12 g/mol + 3 × 16 g/mol = 52 g/mol + 48 g/mol = 100 g/mol

So, n = m/M

= 50 g/100 g/mol

= 0.5 mol

Since From the chemical equation, 1 mole of CaCO₃ produces 1 mole of CaO, then 0.5 × 1 mole of CaCO₃ produces 0.5 × 1 mole of CaO, which is 0.5 mole of CaCO₃ produces 0.5 mole of CaO.

So, we have 0.5 mol of CaO.

We find the mass of CaO from

n =m'/M' where n = number of moles of CaO = 0.5, m' = mass of CaO and M' = molar mass of CaO = 40 g/mol + 16 g/mol = 56 g/mol

So, m' = nM'

= 0.5 mol × 56 g/mol

= 28 g

You might be interested in

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

The most well-known biogenic d-elements

erma4kov [3.2K]

Answer:

i think interstellar dust......

4 0

3 years ago

What happens when magma cools during the rock cycle

Mumz [18]

Answer:

Si las condiciones para que el magma permanezca líquido no perduran, el magma se enfriará y solidificará en una roca ígnea. Una roca que se enfría en el interior de la Tierra se denomina intrusiva o plutónica y su enfriamiento será muy lento, produciendo una estructura cristalina de granos grueso.

Explanation:

7 0

3 years ago

Writing descriptive essay:the best villain Pre-writing

Dominik [7]

Wait so we write the essay?

7 0

3 years ago

Read 2 more answers

The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?

Roman55 [17]

Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

H: 1 g/mole
O: 16 g/mole

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

$moles of water=\frac{3.11 grams*1 mole}{18 gramos}$

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

$heat=\frac{0.1728 moles*40.66 kJ}{1 mole}$

heat= 7.026 kJ

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

7 0

3 years ago