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4 years ago

When iron (II) hydroxide is mixed with phosphoric acid, iron (II) phosphate precipitate results.

Chemistry

1 answer:

Triss [41]4 years ago

3 0

Answer:

a. Fe(OH)2

b. H3PO4

c. 0.01185 moles

Explanation:

The balanced equation of the question will be:

3 Fe(OH)2 + 2 H3PO4 → Fe3(PO4)2 + 6 H2O

a. what is the limiting reactant?

A limiting reactant is a reactant that will be used up in the reaction. We can find it by calculating the stoichiometric ratio. The reactant with the lowest ratio, which will be depleted first, is the limiting reactant. The reaction will stop if any reactant depleted, so the limiting reactant will be the one that determines how many products made. If you add more of a limiting reactant, the reaction will continue.

The molecular mass of Fe(OH)2 is 90g/mol while the molecular mass of H3PO4 is 98 g/mol. The number of each reactant in moles will be

Fe(OH)2 = 3.20g/ (90g/mol) =0.03555 mol

H3PO4 = 2.50g/ (98g/mol) = 0.02551 mol

The actual ratio of Fe(OH)2 : H3PO4 will be:

0.03555 mol/ 0.02551 mol= 1.393 mol/ 1 mol

The stoichiometric coefficient of Fe(OH)2 is 3 while H3PO4 is 2. Mean you need 3 moles Fe(OH)2 of to for the reaction every 2 moles of H3PO4.

The stoichiometric ratio will be: 3 mol/ 2 mol= 1.5 mol/ 1 mol

Since the actual ratio (1.393) is lower than the stoichiometric ratio(1.5) that means we have less Fe(OH)2 than H3PO4, thus the limiting reactant is the Fe(OH)2.

b. What is the reactant in excess?

The excess reactant is the reactant that will have to remain after the reaction stop. It's the opposite of the limiting reactant. Since we still have remains, adding excess reactant won't continue the reaction.

We already know that Fe(OH)2 is the limiting reactant, then H3PO4 will be the excess reactant. The number of reactant remaining will be:0.02551 mol -0.03555 mol/ (1.5 mol -1 mol)=0.00181 mol

c. What is the theoretical yield

Theoretical yield is the predicted amount of product made if the reaction happens with a 100% yield. To find the yield, we need to know how much reactant used and then multiply it with the stoichiometric coefficient of the product. If Fe(OH)2 the limiting reactant then it means all 0.03555 mol of it will be used. The amount of iron (iii) phosphate made will be:

0.03555 mol * 1/3= 0.01185 moles

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Part 2 : No, this reaction will not take place.

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4 years ago

16.78 A buffer is prepared by mixing 525 mL of 0.50 M formic acid, HCHO2, and 475 mL of 0.50 M sodium for- mate, NaCHO2. Calcula

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Answer: Before addition of HCl, the pH is 3.70 and after addition of HCl, the pH is 3.64 .

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$pH=pKa+log(\frac{base}{acid})$

Let's calculate the moles of acid and base(salt) present in the original buffer.

mL are converted to L and then multiplied by molarity to get the moles.

$525mL(\frac{1L}{1000mL})(\frac{0.50molHCHO_2}{1L})$

= $0.2625molHCHO_2$

$475mL(\frac{1L}{1000mL})(\frac{0.50molNaCHO_2}{1L})$

= $0.2375molNaCHO_2$

Total volume of the buffer solution = 0.525 L + 0.475 L = 1.00 L

Since, the total volume is 1.00 L, concentration of base will be 0.2375 M and the concentration of acid will be 0.2625 M.

pKa for formic acid is 3.74. Let's plug in the values in the equation and calculate the pH of the original buffer.

$pH=3.74+log(\frac{0.2375}{0.2625})$

pH = 3.74 - 0.04

pH = 3.70

Now, we add 8.6 mL of 0.15 M HCl acid to 85 mL of the buffer. Let's calculate the moles of acid and base in 85 mL of the buffer.

$85mL(\frac{1L}{1000mL})(\frac{0.2625molHCHO_2}{1L})$

= $0.0223molHCHO_2$

$85mL(\frac{1L}{1000mL})(\frac{0.2375molNaCHO_2}{1L})$

= $0.0202molNaCHO_2$

Now, let's calculate the moles of HCl added to the buffer.

$8.6mL(\frac{1L}{1000mL})(\frac{0.15molHCl}{1L})$

= 0.00129 mol HCl

This added HCl reacts with base(sodium formate) and formic acid is produced.

So, 0.00129 moles of HCl will react with 0.00129 moles of sodium formate to produce 0.00129 moles of formic acid. We can write formate ion in place of sodium formate and hydrogen ion in place of HCl. The equation would be:

$H^++CHO_2^-\rightarrow HCHO_2$

moles of base after reaction with HCl = 0.0202 mol - 0.00129 mol = 0.01891 mol

moles of acid after addition of HCl = 0.0223 mol + 0.00129 mol = 0.02359 mol

Let's plug in the values again in the Handerson equation and calculate the pH:

$pH=3.74+log(\frac{0.01891}{0.02359})$

pH = 3.74 - 0.096

pH = 3.64

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