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marin [14]1 year ago

8 0

The molarity of the acid given the data from the question is 0.30 M

<h3>Balanced equation </h3>

2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, HNO₃ (nA) = 2
The mole ratio of the base, Ba(NO₃)₂ (nB) = 1

<h3>How to determine the molarity of the acid</h3>

From the question given above, the following data were obtained:

Volume of acid, HNO₃ (Va) = 39.7 mL
Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
Molarity of acid, HNO₃ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 39.7) / (0.25 × 24) = 2

(Ma × 39.7) / 6 = 2

Cross multiply

Ma × 39.7 = 6 × 2

Ma × 39.7 = 12

Divide both side by 39.7

Ma = 12 / 39.7

Ma = 0.30 M

Learn more about titration:

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

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<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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Given mass of iron = 4.31 g

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$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

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Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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