Answer:
485.76 g of CO₂ can be made by this combustion
Explanation:
Combustion reaction:
2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)
If we only have the amount of butane, we assume the oxygen is the excess reagent.
Ratio is 2:8. Let's make a rule of three:
2 moles of butane can produce 8 moles of dioxide
Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂
We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
I know the answer. It is 2 straight chains. lLOL
Answer:
Actually, The Henderson - Hasselbalch equation allows you to calculate the pH of the buffer by using the pKa of the weak acid and the ratio that exists between the concentrations of the weak cid and conjugate base. The pKa of formic acid is equal to 3.75. In this case, the pH of the solution will be equal to the acid's pKa .
