Answer:
12.1-M requires less volume of stock solution.
Explanation:
Hello there!
In this case, according to the foundations of dilution processes, it turns out convenient to remember that the moles of the solute before and after the dilution remains the same; for that reason we use the following equation:
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
Relating the initial and final molarities and volumes. Now, it is seen that V2 is 10.0 mL and M2 is 1.0 M, which means that if we solve for V1 for 12.1 M and 6.0 M, we obtain the following volumes:
![V_1=\frac{M_2V_2}{M_1} \\\\V_1^{12.1M}=\frac{1.0M*10.0 mL}{12.1M}=0.826mL\\\\V_1^{6.0M}=\frac{1.0M*10.0 mL}{6.0M}=1.67mL](https://tex.z-dn.net/?f=V_1%3D%5Cfrac%7BM_2V_2%7D%7BM_1%7D%20%5C%5C%5C%5CV_1%5E%7B12.1M%7D%3D%5Cfrac%7B1.0M%2A10.0%20mL%7D%7B12.1M%7D%3D0.826mL%5C%5C%5C%5CV_1%5E%7B6.0M%7D%3D%5Cfrac%7B1.0M%2A10.0%20mL%7D%7B6.0M%7D%3D1.67mL)
Therefore, since the 12.1-M solution requires less volume of stock solution, it makes sense to use this one as the initial solution.
Best regards!
Hello!
The chemical equation for this reaction is the following:
2HCl (aq) + Ca(OH)₂(aq) → CaCl₂(aq) + H₂O(l)
To determine the grams of Calcium Chloride formed, we will need to determine the moles of each reactant and divide by the reaction coefficient to know which is the limiting reactant (the one that is in the lowest amount):
![moles HCl= \frac{vHCl*cHCl}{2} = \frac{1,25L*2,50M}{2} =1,5625 molesHCl \\ \\ molesCa(OH)_2=vCa(OH)_2*cCa(OH)_2=1,10L*2,25M \\ moles Ca(OH)_2=2,475 moles Ca(OH)_2](https://tex.z-dn.net/?f=moles%20HCl%3D%20%5Cfrac%7BvHCl%2AcHCl%7D%7B2%7D%20%3D%20%5Cfrac%7B1%2C25L%2A2%2C50M%7D%7B2%7D%20%3D1%2C5625%20molesHCl%20%5C%5C%20%5C%5C%20molesCa%28OH%29_2%3DvCa%28OH%29_2%2AcCa%28OH%29_2%3D1%2C10L%2A2%2C25M%20%5C%5C%20moles%20Ca%28OH%29_2%3D2%2C475%20moles%20Ca%28OH%29_2)
So, the limiting reactant is HCl, so we'll use the following conversion factor to determine the mass of CaCl₂ produced:
![1,25 L HCl* \frac{2,50 mol HCl}{1 L}* \frac{1 mol CaCl_2}{2 mol HCl}* \frac{110,98 g CaCl_2}{1 mol CaCl_2}= 173,41 g CaCl_2](https://tex.z-dn.net/?f=1%2C25%20L%20HCl%2A%20%5Cfrac%7B2%2C50%20mol%20HCl%7D%7B1%20L%7D%2A%20%5Cfrac%7B1%20mol%20CaCl_2%7D%7B2%20mol%20HCl%7D%2A%20%5Cfrac%7B110%2C98%20g%20CaCl_2%7D%7B1%20mol%20CaCl_2%7D%3D%20%20%20173%2C41%20g%20CaCl_2%20)
So,
173,41 grams of CaCl₂ are produced.
Have a nice day!
Answer:
2.48626 x 10^24
Explanation:
We multiple 4.13 by avogadro's number to get that.
2.89 moles of water would be produced by the complete combustion of 44.5 grams of ethanol in the presence of excess oxygen.
- According to this question, the following reaction between ethanol and oxygen is given:
- C₂H₆O(l) + O₂(g) → CO₂(g) + H₂O(l)
- However, this equation is not balanced. The balanced equation is as follows:
- C₂H₆O + 3O₂ → 2CO₂ + 3H₂O
- According to this equation, 1 mole of ethanol produces 3 moles of water.
- We convert 44.5g of ethanol to moles by dividing by its molar mass as follows:
- mole = 44.5g ÷ 46.07g/mol
- mole = 0.966mol of ethanol
- If 1 mole of ethanol produces 3 moles of water.
- 0.966mol of ethanol will produce (0.966 × 3) = 2.89mol
- Therefore, 2.89 moles of water would be produced by the complete combustion of 44.5 grams of ethanol in the presence of excess oxygen.
Learn more at: brainly.com/question/21085277?referrer=searchResults