Chemistry => Stoichiometry => Limiting reactant
The maximum number of molecules that can be formed will depend on the limiting reactant. The limiting reactant corresponds to the reactant that produces the least amount of product, or in other words, the one that is completely consumed in the reaction.
To find the limiting reactant we are going to divide the moles of each reactant by the stoichiometric coefficients of the balanced equation, the reactant with the lowest ratio will be the limiting reactant.
The balanced equation for this reaction will be:
![Xe+2F_2\rightarrow XeF_4](https://tex.z-dn.net/?f=Xe%2B2F_2%5Crightarrow%20XeF_4)
We have, according to the image the following number of molecules:
Xe=2molecules
F2=4molecules
The limiting reactant will be:
![\begin{gathered} Xe\rightarrow\frac{2}{1}=2 \\ \\ F_2\rightarrow\frac{4}{2}=2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20Xe%5Crightarrow%5Cfrac%7B2%7D%7B1%7D%3D2%20%5C%5C%20%20%5C%5C%20F_2%5Crightarrow%5Cfrac%7B4%7D%7B2%7D%3D2%20%5Cend%7Bgathered%7D)
Both reactants are limiting reactants. We can take any reactant and we will have the same number of molecules of XeF4 formed. The ratio Xe to XeF4 is 1/1, so the molecules of XeF4 that can be formed will be:
![molXeF_4=2molXe\times\frac{1molXeF_4}{1molXe}=2molXeF_4](https://tex.z-dn.net/?f=molXeF_4%3D2molXe%5Ctimes%5Cfrac%7B1molXeF_4%7D%7B1molXe%7D%3D2molXeF_4)
Answer: 2 XeF4 molecules
Answer:
![\boxed {\boxed {\sf 0.055 \ mol \ CO_2}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%200.055%20%5C%20mol%20%5C%20CO_2%7D%7D)
Explanation:
To convert form grams to moles, the molar mass must be used. This is the mass (in grams) in 1 mole of a substance.
We can use the values on the Periodic Table. First, find the molar masses of the individual elements: carbon and oxygen.
- C: 12.011 g/mol
- O: 15.999 g/mol
Check for subscripts. The subscript of 2 after O means there are 2 oxygen atoms, so we have to multiply oxygen's molar mass by 2 before adding.
- O₂: 2* (15.999 g/mol)=31.998 g/mol
- CO₂: 12.011 g/mol + 31.998 g/mol =40.009 g/mol
Use the molar mass as a ratio.
![\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}](https://tex.z-dn.net/?f=%5Cfrac%20%7B44.009%20%5C%20g%5C%20CO_2%7D%7B%201%20%5C%20mol%20%5C%20CO_2%7D)
Multiply by the given number of grams.
![2.4 \ g \ CO_2 *\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}](https://tex.z-dn.net/?f=2.4%20%5C%20g%20%5C%20CO_2%20%2A%5Cfrac%20%7B44.009%20%5C%20g%5C%20CO_2%7D%7B%201%20%5C%20mol%20%5C%20CO_2%7D)
Flip the fraction so the grams of carbon dioxide cancel.
![2.4 \ g \ CO_2 *\frac { 1 \ mol \ CO_2}{44.009 \ g\ CO_2}](https://tex.z-dn.net/?f=2.4%20%5C%20g%20%5C%20CO_2%20%2A%5Cfrac%20%7B%201%20%5C%20mol%20%5C%20CO_2%7D%7B44.009%20%5C%20g%5C%20CO_2%7D)
![2.4 *\frac { 1 \ mol \ CO_2}{44.009}](https://tex.z-dn.net/?f=2.4%20%20%2A%5Cfrac%20%7B%201%20%5C%20mol%20%5C%20CO_2%7D%7B44.009%7D)
![\frac { 2.4 \ mol \ CO_2}{44.009}= 0.0545342998 \ mol \ CO_2](https://tex.z-dn.net/?f=%5Cfrac%20%7B%202.4%20%5C%20mol%20%5C%20CO_2%7D%7B44.009%7D%3D%200.0545342998%20%5C%20mol%20%5C%20CO_2)
The original measurement of grams has 2 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.
The ten thousandth place has a 5, so we round the 4 to a 5.
![0.055 \ mol \ CO_2](https://tex.z-dn.net/?f=0.055%20%5C%20mol%20%5C%20CO_2)
2.4 grams of carbon dioxide is about 0.055 moles.
A substance in a solid phase is relatively rigid, has a definite volume and shape.
<span>The atoms or molecules that comprise a solid are packed close together and are not compressible.<span>Because all solids have some thermal energy, its atoms do vibrate. However, this movement is very small and very rapid, and cannot be observed under ordinary conditions.</span></span>
Answer: I’m ur dad
Explanation:
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