The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.
Further Explanation:
To solve this problem, follow the steps below:
- Write the balanced chemical equation for the given reaction.
- Convert the mass of calcium carbonate into moles.
- Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
- Convert the number of moles of calcium oxide into mass.
Solving the given problem using the steps above:
STEP 1: The balanced chemical equation for the given reaction is:
![CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}](https://tex.z-dn.net/?f=CaCO_%7B3%7D%20%5Crightarrow%20%5C%20CaO%20%5C%20%2B%20%5C%20CO_%7B2%7D)
STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.
![mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}](https://tex.z-dn.net/?f=mol%20%5C%20CaCO_%7B3%7D%20%5C%20%3D%2025%20%5C%20g%20%5C%20CaCO_%7B3%7D%20%5C%20%28%5Cfrac%7B1%20%5C%20mol%20%5C%20CaCO_%7B3%7D%7D%7B100.0869%20%5C%20g%20%5C%20CaCO_%7B3%7D%7D%29%5C%5C%20%5C%5C%5Cboxed%20%7Bmol%20%5C%20CaCO_%7B3%7D%20%5C%20%3D%200.2498%20%5C%20mol%7D)
STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.
For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,
![mol \ CaO \ = 0.2498 \ mol](https://tex.z-dn.net/?f=mol%20%5C%20CaO%20%5C%20%3D%200.2498%20%5C%20mol)
STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.
![mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g](https://tex.z-dn.net/?f=mass%20%5C%20CaO%20%5C%20%3D%200.2498%20%5C%20mol%20%5C%20CaO%20%5C%20%28%5Cfrac%7B56.0774%20%5C%20g%20%5C%20CaO%7D%7B1%20%5C%20mol%20%5C%20CaO%7D%29%5C%5C%20%5C%5Cmass%20%5C%20CaO%20%5C%20%3D%2014.008%20%5C%20g)
Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.
Therefore,
![\boxed {mass \ CaO \ = 14 \ g}](https://tex.z-dn.net/?f=%5Cboxed%20%7Bmass%20%5C%20CaO%20%5C%20%3D%2014%20%5C%20g%7D)
Learn More
- Learn more about stoichiometry brainly.com/question/12979299
- Learn more about mole conversion brainly.com/question/12972204
- Learn more about limiting reactants brainly.com/question/12979491
Keywords: thermal decomposition, stoichiometry
NaBrO3 is the chemical formula for Sodium Bromate.
Answer:
40.94 g
Explanation:
Given data:
Mass of NO₂ = ?
Volume = 20.0 L
Pressure = 110.0 Pka
Temperature = 25°C
Solution:
Pressure = 110.0 KPa (110/101 = 1.1 atm)
Temperature = 25°C (25+273 = 298.15 K)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
n = PV/RT
n = 1.1 atm × 20.0 L / 0.0821 atm.L/ mol.K ×298.15 K
n = 22 / 24.5 /mol
n= 0.89 mol
Mass of NO₂:
Mass = number of moles × molar mass
Mass = 0.89 mol × 46 g/mol
Mass = 40.94 g
Answer:
two north poles and two south poles
Explanation:
A single magnet has a north pole and a south pole. If it is broken into two pieces, then each of the two pieces will have a north pole and a south pole.
No matter how many times or into how many pieces a magnet is broken, the resulting pieces will have two poles each.
Answer:
MOLAR MASS = 32 g/mol
Explanation:
Condition of standard temperature and pressure(STP) are as follow:
Temperature = 273 K
Pressure = 1 atm (or 100000 Pa)
Here atm is atmosphere and Pa is Pascal
STP conditions arte used for measuring gas density and volume using Ideal Gas Law.Here 1 mole of ideal gas occupies 22.4 L of volume.
According toi Ideal Gas Equation :
PV = nRT
where P = pressure, n= number of moles, V = volume ,R= Ideal Gas Constant and T= temperature
![n=\frac{PV}{RT}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D)
From question:
V=280 ml = 0.28 L
P = 1 atm
R=0.08205 L atm/K mol
T=273 K
Putting values in above formula :
![n=\frac{1\times .280}{0.08205\times 273}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1%5Ctimes%20.280%7D%7B0.08205%5Ctimes%20273%7D)
n = 0.0125 moles
Now ![n=\frac{given\ mass}{Molar\ mass}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bgiven%5C%20mass%7D%7BMolar%5C%20mass%7D)
![Molar\ mass=\frac{given\ mass}{n}](https://tex.z-dn.net/?f=%3Cstrong%3EMolar%5C%20mass%3D%5Cfrac%7Bgiven%5C%20mass%7D%7Bn%7D%3C%2Fstrong%3E)
given mass = 0.4 g (given)
![Molar\ mass=\frac{0.4}{0.0125}](https://tex.z-dn.net/?f=%20%3Cstrong%3EMolar%5C%20mass%3D%5Cfrac%7B0.4%7D%7B0.0125%7D%3C%2Fstrong%3E)
On solving we get:
Molar mass = 32 g/mol