QuestionIdentify the limiting reactant and calculate the max mass of from , , and .
Studdy Solution
STEP 1
Assumptions1. The reaction is balanced.
. The reaction goes to completion.
3. The mass of Ru3 is121 g.
4. The mass of CO is39.7 g.
5. The mass of Ag is71.9 g.
STEP 2
First, we need to calculate the molar mass of each reactant and product involved in the reaction.The molar mass of Ru is the sum of the molar mass of Ru and three times the molar mass of I.The molar mass of CO is the sum of the molar mass of C and O.The molar mass of Ag is the molar mass of Ag itself.The molar mass of Ru(C)5 is the sum of the molar mass of Ru and five times the molar mass of CO.
STEP 3
Now, calculate the number of moles of each reactant available in the mixture. We can do this by dividing the given mass of each reactant by its molar mass.
STEP 4
Next, calculate the number of moles of Ru(C) that each reactant can produce if it was the limiting reactant. We can do this by multiplying the number of moles of each reactant by the stoichiometric ratio of Ru(C) to that reactant.
\text{Moles of Ru(C)_ from Ru}_3 = \text{Moles of Ru}_3 \times \frac{1}{1}\text{Moles of Ru(C)_ from CO} = \text{Moles of CO} \times \frac{1}{}\text{Moles of Ru(C)_ from Ag} = \text{Moles of Ag} \times \frac{1}{3}
STEP 5
The reactant that produces the least amount of Ru(C)5 is the limiting reactant. Compare the number of moles of Ru(C)5 that each reactant can produce to identify the limiting reactant.
STEP 6
Once the limiting reactant is identified, calculate the maximum mass of Ru(C)5 that can be produced. We can do this by multiplying the number of moles of Ru(C)5 that the limiting reactant can produce by the molar mass of Ru(C)5.
\text{Mass of Ru(C)_5} = \text{Moles of Ru(C)_5 from limiting reactant} \times \text{molar mass of Ru(C)_5}
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