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soldi70 [24.7K]
3 years ago
11

Upon balancing the equation below, how many moles of sulfuric acid are needed to react completely with 8.4 moles of lithium hydr

oxide?
LiOH + H2SO4 yields Li2SO4 + H2O
a. 2.1 moles
b. 4.2 moles
c. 8.4 moles
d. 16.8 moles
Chemistry
1 answer:
maks197457 [2]3 years ago
7 0
The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
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What drives spontaneous reactions?
KatRina [158]

Answer:

option B

Explanation:

A spontaneous reaction drive favorable when enthalpy is decreasing and the entropy is increasing on the system. If that happens the reaction occurs spontaneously

5 0
3 years ago
Read 2 more answers
At 25 ∘C , the equilibrium partial pressures for the reaction were found to be PA=5.16 bar, PB=5.04 bar, PC=4.11 bar, and PD=4.8
erastova [34]

Answer: 5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)

The expression for equilibrium reaction will be,

K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}

Now put all the given values in this expression, we get the concentration of methane.

K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}

K_p=10.6

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

\Delta G^o=-2.303\times RT\times \log K_c

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^0C=(25+273)K=298 K

K_c = equilibrium constant = 10.6

\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)

\Delta G^o=5850.23J/Kmol

\Delta G^o=5.85kJ/Kmol

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

3 0
3 years ago
Limiting reactants would appreciate the help
Vanyuwa [196]

Answer:

Explanation:

The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.

To identify the limiting reactant ...

- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).

- convert all given reactant values to moles

- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.  

Your Problem:

Given:        3Ba  +  N₂  => Ba₃N₂

               22.6g    4.2g        ?

moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba

moles N₂ =>    4.2g/14.007g/mol= 0.150 mole N₂

Part A: Determining the Limited Reactant

  • Divide each mole value by respective coefficient ... smallest value is Limiting  Reactant.

Barium => 0.165/3 = 0.055  <=> (Limiting Reactant)

Nitrogen => 0.15/1 = 0.15

  • Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.

Part B: Max (theoretical) amount of Ba₃N₂ produced:

<em>Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.     </em>

                   3Ba        +          N₂         =>     Ba₃N₂    (3:1 rxn ratio for Ba:Ba₃N₂)

moles      0.165mole        0.150mole         1/3(0.165)mole = 0.055mole Ba₃N₂

                                                                    = 0.055mol(440g/mol) Ba₃N₂

                                                                    = 24.2 grams Ba₃N₂ (as based

                                                                     upon Barium as Limiting Reactant)

Part C: Excess N₂ remaining after reaction stops:

From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) =  moles of N₂ used.

moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used  

∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.

mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.

                   

5 0
3 years ago
How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04 ​
Free_Kalibri [48]

Answer:

3

Explanation:

the three elements involved in this compound are Li, S, O.

lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."

7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.

6 0
3 years ago
How did changing from a light drizzle to a downpour affect the river and the sediment un it
soldi70 [24.7K]

Answer:

The velocity of the river increased.

There was more erosion in the stream.

The type of sediment that moved changed.

Explanation:

6 0
3 years ago
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