Answer:
It's because they aren't destroyed they're just rearranged differently to create different products. There's like a name for this. It's something like the law of conservation energy. But the reason is mass cannot be created nor can it be destroyed.
Answer:
![m_{H_2O}=0.353gH_2O](https://tex.z-dn.net/?f=m_%7BH_2O%7D%3D0.353gH_2O)
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:
![1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O](https://tex.z-dn.net/?f=1.4gCO_2%2A%5Cfrac%7B1molCO_2%7D%7B44.01gCO_2%7D%2A%5Cfrac%7B1molH_2O%7D%7B1molCO_2%7D%3D0.0318mol%20%20H_2O%5C%5C%5C%5C2.2gKOH%2A%5Cfrac%7B1molKOH%7D%7B56.11gKOH%7D%2A%5Cfrac%7B1molH_2O%7D%7B2molKOH%7D%3D0.0196mol%20%20H_2O)
In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:
![m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O](https://tex.z-dn.net/?f=m_%7BH_2O%7D%3D0.0196molH_2O%2A%5Cfrac%7B18.02gH_2O%7D%7B1molH_2O%7D%3D0.353gH_2O)
Regards!
Answer:
I also think NaOH is the answer since they are all strong electrolytes and NaOH has the highest number
1.2*10^24# atoms of chlorine
Explanation:
Chlorine gas (#Cl_2#) has two atoms of elemental chlorine in a molecule, so:
#1# mol of #Cl_2# have #6*10^23# molecules of #Cl_2#
#1# molecule of #Cl_2# have #2# atoms per molucule
Then #2*6*10^23 = 1.2*10^24# atoms of chlorine in a mol of chlorine gas
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
![\dfrac{80}{17}=4.706](https://tex.z-dn.net/?f=%5Cdfrac%7B80%7D%7B17%7D%3D4.706)
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
![\tt \dfrac{120}{32}=3.75](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B120%7D%7B32%7D%3D3.75)
Mol ratio of reactants(to find limiting reatants) :
![\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B4.706%7D%7B4%7D%5Cdiv%20%5Cdfrac%7B3.75%7D%7B5%7D%3D1.1765%5Cdiv%200.75%5Crightarrow%20O_2~limiting~reactant%28smaller~ratio%29)
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
![\tt \dfrac{6}{5}\times 3.75=4.5](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B6%7D%7B5%7D%5Ctimes%203.75%3D4.5)
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
![\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%](https://tex.z-dn.net/?f=%5Ctt%20%5C%25yield%3D%5Cdfrac%7Bactual%7D%7Btheoretical%7D%5Ctimes%20100%5C%25%5C%5C%5C%5C%5C%25yield%3D%5Cdfrac%7B72.2%7D%7B81%7D%5Ctimes%20100%5C%25%3D89.14%5C%25)