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Zarrin [17]
2 years ago
10

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

hydrogen and 20.0 g of iodine are heated, forming 10.0 g of hydrogen iodide, what mass of hydrogen remains unreacted? A. 10.0 g hydrogen remains B. 10.9 g hydrogen remains C. 15.0 g hydrogen remains D. 19.9 g hydrogen remains.
Chemistry
1 answer:
Kaylis [27]2 years ago
5 0

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

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D = mass / volume

d = 100 g / 100 mL

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3 years ago
How do polar and nonpolar covalent bonds differ?
3241004551 [841]

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Explanation:

7 0
3 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

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As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

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\Delta S=\frac{\Delta H_{freezing}}{T_m}

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4 0
3 years ago
HI(aq)+NaOH(aq)→ <br> what the final balanced chemical equation with the phases included
julia-pushkina [17]

Answer:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

HI(aq)+NaOH(aq)\rightarrow NaI+H_2O

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Which is also balanced as the number of atoms of all the elements is the same at both sides.

Best regards!

7 0
3 years ago
Please help! will mark brainliest
Step2247 [10]

Answer:

add me!!!

Explanation:

my insta is: becomewhatyouwant2inlife

3 0
2 years ago
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