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myrzilka
20 hours ago
10

If 4.35 g of phosphoric acid are added to 5.25g of KOH, what is the percent yield of the reaction if only 3.15g of potassium pho

sphate is isolated?
Chemistry
1 answer:
lorasvet [946]19 hours ago
5 0
We are provided with
4.35 g of phosphoric acid
5.25 g of KOH
3.15 g of K3PO4 produced

The reaction formula is
H3PO4 + 3KOH => K3PO4 + 3H2O

Initially, convert the given masses into moles.
Then, identify the limiting reactant. Afterward, calculate the maximum K3PO4 that can be generated from the limiting reactant.
Finally, compute the percent yield by dividing the actual yield by the theoretical yield.
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A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio
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Answer:

The yield percentage of H_2CO_3 is 24.44%

Explanation:

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12 days ago
Which change of state is shown in the model?
Alekssandra [962]

I think the state change illustrated in the diagram is deposition. 
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9 days ago
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A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
KiRa [965]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

Por lo tanto,

0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

3 0
14 days ago
How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?
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Response:

9.9 ml of 0.200M NH₄OH(aq)

Reasoning:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?

1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution

1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)

=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters

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