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Answer : The reagent present in excess and remains unreacted is,
Solution : Given,
Moles of = 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Answer: The correct answer is: " 0.54 g " .
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Explanation:
Note that "hydrogen gas" is:
H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;
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The molecular weight of "H" is: 1.00794 g ;
(From the Periodic Table of Elements).
So, the molecular weight of: H₂ (g) is:
" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;
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Note the conversion for a gas at STP:
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1 mol of a gas = 22.4 L gas;
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i.e. " 1 mol / 22.4 L " ;
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So: " 5.7 L H₂ (g)
The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;
The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;
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and we are left with:
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[5.7 * 2.104588 g ] / 22.4 = ? g ;
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→ [ 11.9961516 g ] / 22.4 =
0.53554248214 g ;l
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We round this value to: " 0.54 g " ;
→ since "5.7 L " has 2 (two) significant figures;
22.4 is an exact number conversion;
and "5.7 L" has fewer significant figures than:
" 2.104588 " ; or: " 1.00794 " .
→ as such: We round to "2 (two) significant figures."
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Hope this is helpful. Wishing you the best in your academic endeavors!
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