Answer:
C:
Explanation:
we can use the molarity equation
so to find M we plug in what we know, which is 6 moles of NaCl and 2 L of water, which gives us:
What volume of hydrogen iodide is produced when 118 liters of hydrogen gas react according to the following reaction? (All gases
1 answer:
Answer:
Explanation:
Hello,
In this case, for the given balanced chemical reaction:
Thus, since hydrogen and hydrogen iodide are in a 1:2 mole ratio, we can easily compute the yielded volume as shown below:
Thus, is possible, due to the Avogadro's law which allows to relate moles and volume by a directly proportional relationship.
Regards.
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Answer:
1.327 g Ag₂CrO₄
Explanation:
The reaction that takes place is:
- 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)
First we need to <em>identify the limiting reactant</em>:
We have:
- 0.20 M * 50.0 mL = 10 mmol of AgNO₃
- 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄
If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.
Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:
- 4 mmol K₂CrO₄ *
= 1326.92 mg Ag₂CrO₄
- 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄