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liberstina [14]
4 years ago
14

Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (

Use the lowest possible coefficients. Include states of matter under the SATP conditions in your answer.)
How many mol H2SO4 is produced from 11.5 moles of SO2?

How many O2 is needed?
Chemistry
1 answer:
Nesterboy [21]4 years ago
7 0

Balanced chemical equation for the reaction is:

2SO_{2} (g) + O_{2} (g)+ 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

Moles of H_{2} S_{} O_{4} formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of SO_{2} = 11.5 moles

moles of H_{2} S_{} O_{4} = ?

Moles of O_{2} needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2SO_{2}(g) + O_{2} (g) + 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of SO_{2}, x moles of H_{2} S_{} O_{4} is formed

\frac{1}{2}  =\frac{x}{11.5}

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1  mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

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How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid? Unbalanced equatio
kap26 [50]
The balanced reaction that describes the reaction between sulfuric and sodium hydroxide to produce sodium sulfate and water is expressed <span>H2SO4 + 2NaOH → 2H2O + Na2SO4. When 65.5 grams of sodium hydroxide reacts with excess sulfur, we need first to convert the mass to moles and multiply by stoich ratio 2/2 or 1. Hence the moles is multiplied to the molar mass of water. The final answer is 29.475 grams water.</span>
8 0
4 years ago
HELPPPP PLEASE ASAP!!!
irina [24]
Answer:

230
     Th
  90

Explanation:

The left side shows the element uranium with mass number 234 (superscript to the left of ths symbol) and atomic mass 92 (subscript to the left of the symbol).

So,  you have to find the mass number and atomic numbers missing on the right side which added to those shown for the helium atom reach 234 and 92.

This the the mathematics:

1) Mass number of U - mass number of He = 234 - 4 = 230 ⇒ the mass number of the missing element is 230.

2) atomic number of U - atomic number of He = 92 - 2 = 90 ⇒ the atomic number of the missing element is 90.

Now, you use the atomic mass to identify the element. You do it by simply searching in a periodic table. There you find that the atomic number 90 corresponds to thorium, whose symbol is Th.

Now you have that the product that completes the right side of the decay equation is:

230
     Th
  90


4 0
4 years ago
Tricky please help !!!!!!!!
Illusion [34]
A, B, and C because they all equal 5.6 cm and it’s not close to 6.
5 0
3 years ago
Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(a
Alchen [17]
E^{o}cell =  E^{o} (Ag+/Ag) -  E^{o}( Sn2+/Sn)
where, E^{o} (Ag+/Ag) = std. reduction potential of Ag+ = 0.7994 v
and Sn2+/Sn = std. reduction potential of Sn2+ = -0.14 v

Thus, E^{o}cell = 0.7994v - (-0.14v) = 0.9394 v

Now, ΔG^{o} = -nFE^{0},
where, n = number of electrons = 2
F = Faraday's constant = 96500 C
∴ΔG^{o} = 2 X 96500 X 0.9394 = -1.18 X 10^{5} J/mol

Now, using Nernst's Equation we have,
[tex]E_{cell} = 0.9394 - \frac{2.303X298}{2X96500}log \frac{0.0115}{ 3.5^{2} }
E_{cell} = 0.9765 v

Finally, ΔG = -nFE = -2 X 96500 X 0.9765 = -1.88 X 10^{5} J/mol

3 0
3 years ago
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What are the products of a neutralization reaction? salt and carbon dioxide carbon dioxide and water water and salt oil and wate
Mrac [35]
Water and salt are the products of a neutralization reaction.
8 0
4 years ago
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