How Many Carbon Atoms are in 4.00 G of Butane
Butane, or C4H10, is a hydrocarbon that contains four carbon atoms. It is a gas at room temperature and pressure, and it is often used as a fuel for lighters and camp stoves. When burned, butane produces carbon dioxide and water vapor.
Butane is an organic compound with the chemical formula C4H10. This means that there are four carbon atoms and ten hydrogen atoms in each molecule of butane. Thus, in 4.00 grams of butane, there would be a total of 40,000 carbon atoms.
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How Many Carbon Atoms are in 1G of Butane?
One gram of butane contains 30.0664 grams of carbon atoms. This is because butane is a hydrocarbon, and all hydrocarbons are made up of hydrogen and carbon atoms. The atomic weight of carbon is 12.0107, so one gram of butane contains 30.0664 grams of carbon atoms (12.0107 x 30.0664 = 360.7832).
How Many Carbon Atoms are in 2.5 G of Butane?
The butane molecule is made up of four carbon atoms and ten hydrogen atoms, for a total atomic mass of 58.12 amu. There are Avogadro’s number of molecules in 2.5 grams of butane, so we can calculate the number of carbon atoms by multiplying Avogadro’s number by the molar mass of carbon (12.0107 amu):
Does Butane Have 4 Carbon Atoms?
Yes, butane has 4 carbon atoms.
How Many Carbon Atoms are in 1G?
There are 12.0107 grams of carbon in 1 mole, and there are Avogadro’s number of carbon atoms in 1 mole. This means that there are approximately 0.012 kilograms of carbon in 1 gram, or 12g/1000g=12000mg/1g. Therefore, there are 12000mg/1g*(6.022*10^23)/(12.0107)=6.0 x 10^23 atoms of carbon in 1 gram.
How to Find the Number of Atoms in C4H10 (Butane)
How Many Moles of Nh3 Can Be Produced
When it comes to the production of moles of NH3, it is important to know the process that is being used. In order to produce one mole of NH3, three moles of hydrogen are needed. This can be accomplished in a number of ways, but the most common method is through the Haber-Bosch process.
This industrial process uses nitrogen and hydrogen gas as reactants in order to produce ammonia. The equation for this reaction is: N2 + 3H2 –> 2NH3
In order to determine how many moles of NH3 can be produced, we need to know the amount of reactants that are available.
It is assumed that an infinite amount of nitrogen gas is available. However, there is only a finite amount of hydrogen gas. So, if we have X moles of hydrogen gas, we can produce 0.5X moles of NH3.
Keep in mind, however, that the Haber-Bosch process has an 80-90% yield. So not allreactants will produce product (in this case, NH3). But assuming 100% yield,.
Ifwe have 12gof H2available,, thenwe could potentiallyproduce 24g or 2molesofNH3224g
But again,, usingthe morerealisticyield(80%-9090%),thenthesame12go fH22wouldonlygiveus between0 . 96and1 .
How Many Moles of Nh3 Can Be Produced from 24.0 Mol of H2 And Excess N2 ?
In order to answer this question, we must first understand what a mole is. A mole is a unit of measurement that refers to the number of atoms or molecules in a given sample. In this case, we are talking about moles of NH3, which means that we are looking at the number of nitrogen and hydrogen atoms in NH3.
So, how many moles of NH3 can be produced from 24.0 mol of H2 and excess N2? Well, according to the equation for the production of NH3, we need 3 moles of H2 for every 1 mole of NH3. This means that we would need 36 moles of H2 in order to produce 12 moles of NH3.
However, since we only have 24 mol of H2, we would not be able to produce that much NH3. In fact, the maximum amount of NH3 that could be produced from 24 mol H2 would be 8 moles (given an excess supply of N2).
How Many Moles of Nh3 Can Be Produced from 18.0 Mol of H2 And Excess N2 ?
In order to answer the question of how many moles of NH3 can be produced from 18.0 mol of H2 and excess N2, we must first understand what a mole is. A mole is simply a unit used to measure the amount of a substance that contains an Avogadro’s number of particles. The Avogadro’s number is 6.02 x 10^23 particles/mole.
So, one mole of any substance contains 6.02 x 10^23 particles of that substance. With that said, let’s get back to the original question at hand.
In order to produce NH3 (ammonia), we need both nitrogen and hydrogen gas.
The chemical equation for this reaction is: N2 + 3H2 –> 2NH3. As we can see from the equation, it takes two moles of nitrogen gas to react with every three moles of hydrogen gas in order to produce two moles of ammonia gas. Therefore, if we have 18 mol of hydrogen gas, we would need 12 mol (6 x 2) or nitrogen gas in order to completely react all the hydrogen and produce an equal amount of ammonia gas.
So far we have determined that in order to produce 18 mol NH3, we would need 12mol N2 and 18mol H2 . However, our problem states that we only have 18mol H2 and “excess” N2 . This means that there is more than enough nitrogen present for the desired reaction to take place–we don’t have to worry about running out before all the hydrogen has been used up!
In fact, with excess nitrogen around, it is likely that not all the hydrogen will be used up and some unreacted nitrogen will remain as well (although less than what was originally present).
To sum it up: In order to produce 18 mol NH3 from H2 and N2 , you would need 18 mol H2 and 12 mol N2 , but because there is excess N2 , not all the H2 will be used up and some unreacted N2 will remain as well.
How Many Atoms are in Carbon
This question can be a bit tricky to answer because it really depends on what you mean by “carbon.” If you are referring to the element carbon, then there are six protons in the nucleus and, therefore, six electrons surrounding the nucleus. However, if you are asking how many atoms of carbon there are in 12 grams of carbon-12 (which is an isotope of carbon), then there are Avogadro’s number of atoms, or 6.022 x 10^23 atoms.
Conclusion
Butane is a hydrocarbon gas that contains four carbon atoms. It has a molecular weight of 58.12 and a boiling point of -0.5 degrees Celsius. Butane is used as a fuel for many things, such as lighters and grills.
It is also used to make other chemicals.