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castortr0y [4]
3 years ago
9

if 334.6 g of phosphoric acid is reacted with excess potassium hydroxide. the final mass K3PO4 produced is found to be 248g. wha

t is the percent yield?
Chemistry
1 answer:
Orlov [11]3 years ago
3 0

Answer:

                     %age Yield  =  34.21 %

Explanation:

                   The balance chemical equation for the decomposition of KClO₃ is as follow;

                            3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O

Step 1: Calculate moles of H₃PO₄ as;

Moles = Mass / M/Mass

Moles = 334.6 g / 97.99 g/mol

Moles = 3.414 moles

Step 2: Find moles of K₃PO₄ as;

According to equation,

                 1 moles of H₃PO₄ produces  =  1 moles of K₃PO₄

So,

              3.414 moles of H₃PO₄ will produce  =  X moles of K₃PO₄

Solving for X,

                      X = 1 mol × 3.414 mol / 1 mol

                      X = 3.414 mol of K₃PO₄

Step 3: Calculate Theoretical yield of K₃PO₄ as,

Mass = Moles × M.Mass

Mass = 3.414 mol × 212.26 g/mol

Mass = 724.79 g of K₃PO₄

Also,

%age Yield  =  Actual Yield / Theoretical Yield × 100

%age Yield  =  248 g / 724.79 × 100

%age Yield  =  34.21 %

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How many mL of a 6 M NaOH stock solution is needed in order to prepare 500 mL of a 0.2 M NaOH solution?
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Answer:

The right answer is "16.67 mL".

Explanation:

Given:

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M_1=6 \ M

M_2=0.2 \ M

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V_1=V \ mL

V_2=500 \ mL

As we know, the equation,

⇒ M_1V_1=M_2V_2

On putting the values, we get

⇒ 6\times V=0.2\times 500

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0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(
Studentka2010 [4]

Answer : The concentration of Fe^{3+} at equilibrium is 0 M.

Solution :  Given,

Concentration of Fe^{3+} = 0.0200 M

Concentration of C_2O_4^{2-} = 1.00 M

The given equilibrium reaction is,

                            Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

The expression of K_c will be,

K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}

1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}

By solving the term, we get:

x=0.02M

Concentration of Fe^{3+} at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of Fe^{3+} at equilibrium is 0 M.

4 0
3 years ago
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