Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma
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Answer:
(A) 0.129 M
(B) 0.237 M
Explanation:
(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:
- 2HA + Ba(OH)₂ → BaA₂ + 2H₂O
Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).
We <u>convert mass of phthalate to moles</u>, using its molar mass:
- 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol
Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:
- 9.27 mmol HA *
= 6.64 mmol Ba(OH)₂
Finally we calculate the molarity of the Ba(OH)₂ solution:
- 6.64 mmol / 35.8 mL = 0.129 M
(B) The reaction between Ba(OH)₂ and HCl is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
So<u> the moles of HCl that reacted </u>are:
- 17.1 mL * 0.129 M *
= 4.41 mmol HCl
And the <u>molarity of the HCl solution is</u>:
- 4.41 mmol / 18.6 mL = 0.237 M
The correct question is as follows:
How do you convert from grams to moles of a substance
1. Divide by the molar mass
2. Subtract the molar mass
3. Add the molar mass
4. Multiple by the molar mass
Answer: In order to convert from grams to moles of a substance divide by the molar mass.
Explanation:
The number of moles of a substance is the mass of substance in grams divided by its molar mass.
The formula to calculate moles is as follows.
This means that grams are converted to moles when grams is divided by molar mass.
Thus, we can conclude that in order to convert from grams to moles of a substance divide by the molar mass.
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.