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klasskru [66]
3 years ago
8

Uranium can be isolated from its Ores by dissolving it as UO2(NO3)2, then separating If as solid UO2(C2O4).3H2O. Addition of 0.4

031 gram of Sodium Oxalate, NaC2O4,to a solution containing 1.481 gram of uranyl nitrate,UO2(NO2)2, yields 1.073 gram of solid UO2(C2O4).3H2O
Na2C2O4 + UO2(NO3)2 3H2O ? UO2(C2O4).3H2O + 2NaNO3
determine the limiting reactant and the percent yield of this reaction
Chemistry
2 answers:
Arada [10]3 years ago
6 0

Answer:

The limiting reactant is NaC₂O₄ and the yield of this reaction is 69.52%.

Explanation:

NaC₂O₄ + UO₂(NO₃)₂ + 3H₂O → UO₂(C₂O₄).3H₂O + 2NaNO₃

m (NaC₂O₄) = 0.4031 g  MW = 134 g/mol ∴ n = 3.0 mmol

m (UO₂(NO₃)₂) = 1.481 g MW = 396.05 g/mol ∴ n = 3.74 mmol

m (UO₂(C₂O₄).3H₂O) = 1.073 g MW = 412.094 g/mol ∴ n = 2.60 mmol

The limiting reactant is NaC₂O₄

The yield can be given by (2.60/3.74).100% = 69.52%

andriy [413]3 years ago
5 0

Answer:

Limiting reactant: Na₂C₂O₄

Percent yield: 86.53%

Explanation:

Let's consider the following reaction.

Na₂C₂O₄ + UO₂(NO₃)₂ + 3H₂O → UO₂(C₂O₄).3H₂O + 2NaNO₃

The molar mass of Na₂C₂O₄ is 134.0 g/mol and the molar mass of UO₂(NO₃)₂ is 394.0 g/mol. In the balanced equation, there is 1 mole of each one.

The theoretical mass ratio of UO₂(NO₃)₂ to Na₂C₂O₄ is 394.0/134.0 = 2.940/1.

The experimental mass ratio of UO₂(NO₃)₂ to Na₂C₂O₄ is 1.481/0.4031 = 3.674/1.

Comparing the theoretical and the experimental mass ratios, we can see that the reactant in excess is UO₂(NO₃)₂ and the limiting reactant is Na₂C₂O₄.

We will use the limiting reactant to calculate the theoretical yield of UO₂(C₂O₄).3H₂O. The molar mass of UO₂(C₂O₄).3H₂O is 412.1 g/mol and the mass ratio of Na₂C₂O₄ to UO₂(C₂O₄).3H₂O is 134.0g / 412.1g.

The theoretical yield of UO₂(C₂O₄).3H₂O is:

0.4031 g Na₂C₂O₄ × (412.1 g UO₂(C₂O₄).3H₂O / 134.0 g Na₂C₂O₄) = 1.240 g UO₂(C₂O₄).3H₂O

The percent yield is:

(1.073 g / 1.240 g) × 100% = 86.53%

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A.) 4.0

Explanation:

The general equilibrium expression looks like this:

K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }

In this expression,

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-----> uppercase letters = molarity

-----> lowercase letters = balanced equation coefficients

In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.

K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }                                       <----- Equilibrium expression

K = \frac{[2 M] [2 M]}{[1 M] [1 M] }                                     <----- Insert molarities

K = \frac{4}{1  }                                                <----- Multiply

K = 4                                                <----- Divide

6 0
2 years ago
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4 0
3 years ago
Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p
Tanzania [10]

Answer :  The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)

The expression for rate of reaction :

\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}

\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}

\text{Rate of disappearance of }F=-\frac{d[F]}{dt}

\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}

\text{Rate of formation of }H=+\frac{d[H]}{dt}

\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}

Given:

-\frac{d[D]}{dt}=0.18mol/L.s

As,  

-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s

and,

+\frac{d[H]}{dt}=2\times 0.18mol/L.s

+\frac{d[H]}{dt}=0.36mol/L.s

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0
3 years ago
Which statement is correct about molarity and volume?
Oksana_A [137]

Answer:

Explanation:

If the choices are:

A. Molarity is defined as moles of solute per liter of solvent.

B. % by mass is defined as grams of solute per 100 g of solvent.

C. % by volume is defined as grams of solute per 100 L of solution.

D. Molarity is defined as moles of solute per liter of solution.

E. All of the above.

then the ans is E

7 0
3 years ago
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