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Fofino [41]
2 years ago
10

2Fe(s) +3H2SO4(aq) →Fe2(SO4)3(aq) +3H2(g)When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the percent y

ield if 5.40 g of "hydrogen gas" are collected?
Chemistry
1 answer:
Elden [556K]2 years ago
7 0

Answer:

1040%

Explanation:

To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:

Percent yield = Actual yield (5.40g) / Theoretical yield * 100

<em>Moles Fe -Molar mass: 55.845g/mol-:</em>

10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.

For a complete reaction of these moles there are necessaries:

0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.

As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>

The moles of H2 produced are:

0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2

The mass is:

0.277 moles H2 * (2.016g/mol) = 0.558g H2

Percent yield is:

5.40g / 0.558g * 100 = 1040%

It is possible the experiment wasn't performed correctly

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The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t
Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

6 0
3 years ago
Why does the melting of the polar ice caps facilitate the absorption of CO2 by ocean waters?​
Korvikt [17]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────⊰

Answer:

The melting of the caps increases the concentration of CO2 in the water, making its absorption faster.

⊱┈────────────────────────⊰

Explanation:

<h2><u>CO2 absorption</u><u> </u></h2>

The polar caps of our planet have in their composition several elements, among them the CO2 that is absorbed by the atmosphere. Cold waters, which are present in the Arctic, have an easier time absorbing CO2 compared to other waters.

When glaciers melt, the CO2 that is present in the mixture is dissolved in the ocean, increasing its concentration. The cold waters that came from the ice caps increase the absorption of CO2.

Learn more about natural methods remove CO2 from the atmosphere in:

brainly.com/question/14323197

#SPJ2

4 0
1 year ago
In a sample of solid ba(no3)2 the ratio of barium ions to nitrate ions is
user100 [1]
<span>In a sample of solid Ba(NO3)2 the ratio of barium ions to nitrate ions is would be one is to 2 or 1:2. Barium ion has a formal charge of positive two which means that it needs two ions which has a formal charge of negative one or 1 ion with the formal charge of negative two. However, for this case, it is bonded to a nitrate ion which has a formal charge of negative one. Therefore, it needs two nitrate ions so that for every 1 atom of barium ion, we need two ions of nitrate ions.</span>
3 0
3 years ago
rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
SVEN [57.7K]

Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

7 0
3 years ago
Read 2 more answers
What is the difference between an enzyme and a catalyst?
stealth61 [152]
Catalyst- chemical that increases the rate of a chemical reaction without itself being changed from the reaction. Enzyme- substance produced by a living organism that acts as a catalyst to bring about a particular biochemical reaction. Hope it helps.
6 0
3 years ago
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