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3 years ago

PLS HELP MEEEE I REALLY NEED HELP

Chemistry

1 answer:

yawa3891 [41]3 years ago

3 0

Answer:

the answeer is c.

Explanation:

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Carbon reacts with steam (h2o) at high temperatures to produce hydrogen and carbon monoxide. if 2 mol of carbon are exposed to 3

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The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
stoichiometry of C to H₂O is 1:1
1 mol of C reacts with 1 mol of H₂O
we need to find which is the limiting reactant
2 mol of C and 3.1 mol of H₂O
therefore C is the limiting reactant and H₂O is in excess.
stoichiometry of C to H₂ is 1:1
then number of H₂ moles formed are equal to C moles reacted
number of H₂ moles formed = 2 mol

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3 years ago

The atmosphere's variable gases that most influence the greenhouse effect are _____. nitrogen and oxygen argon and carbon dioxid

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2 years ago

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Salicylic acid (C7H6O3) reacts with acetic anhydride (C4H6O3) to form acetylsalicylic acid (C9H8O4).

zavuch27 [327]

Answer : The correct option is, (B) Salicylic acid

Solution :

First we have to calculate the moles of salicylic acid and acetic anhydride.

$\text{Moles of }C_7H_6O_3=\frac{\text{Mass of }C_7H_6O_3}{\text{Molar mass of }C_7H_6O_3}=\frac{70g}{138.121g/mole}=0.507moles$

$\text{Moles of }C_4H_6O_3=\frac{\text{Mass of }C_4H_6O_3}{\text{Molar mass of }C_4H_6O_3}=\frac{80g}{102.09g/mole}=0.783moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_7H_6O_3(aq)+C_4H_6O_3(aq)\rightarrow 2C_9H_8O_4(aq)+H_2O(l)$

From the balanced reaction we conclude that

As, 2 moles of salicylic acid react with 1 mole of acetic anhydride

So, 0.507 moles of salicylic acid react with $\frac{0.507}{2}=0.2535$ mole of acetic anhydride

The excess of acetic anhydride = 0.783 - 0.2535 = 0.5295 moles

That means the in the given balanced reaction, salicylic acid is a limiting reagent because it limits the formation of products and acetic anhydride is an excess reagent.

Hence, the limiting reagent is, salicylic acid.

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3 years ago

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Answer:

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Explanation:

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