Question

Titration of a 0.7439 g sample of impure Na2B4O7(borax) required 31 64 mL of O. 108. M HCl for reaction. In terms of B2O3, percentage is
B40+ 2H30+ + 3H20 + 4H3B03​

Answer 1

Therefore, the percentage of $B_2O_3$ in the reaction product is 27.81%.

The balanced chemical equation for the reaction between borax $(Na_2B_4O_7)$ and hydrochloric acid (HCl) is:

$Na_2B_4O_7 + 2HCl + 5H_2O \longrightarrow 4H_3BO_3 + 2NaCl$

From the equation, we can see that 1 mole of borax reacts with 2 moles of HCl to produce 4 moles of boric acid $(H_3BO_3)$.

The molar mass of borax is 201.22 g/mol, which means that the number of moles of borax in 0.7439 g sample can be calculated as follows:

0.7439 g / 201.22 g/mol = 0.0037007 mol of borax

The volume of HCl used in the reaction is 31.64 mL, or 0.03164 L.

The concentration of the HCl solution is 0.108 M, which means that the number of moles of HCl used in the reaction can be calculated as follows:

0.108 mol/L x 0.03164 L = 0.003422 mol of HCl

According to the balanced chemical equation, 1 mole of borax reacts with 2 moles of HCl.

Therefore, the number of moles of borax that reacted can be calculated as follows:

0.003422 mol of HCl / 2 = 0.001711 mol of borax

The number of moles of $B_2O_3$ in the reaction can be calculated from the number of moles of borax using the stoichiometry of the balanced chemical equation. From the equation, we can see that 1 mole of borax contains 2 moles of  $B_2O_3$ , which means that the number of moles of  $B_2O_3$  in the sample can be calculated as follows:

0.001711 mol of borax x 2 mol of B2O3/mol of borax = 0.003422 mol of  $B_2O_3$

The mass of  $B_2O_3$  in the sample can be calculated from the number of moles of  $B_2O_3$  using its molar mass, which is 69.62 g/mol:

0.003422 mol of  $B_2O_3$ x 69.62 g/mol = 0.238 g of B2O3

Therefore, the amount of  $B_2O_3$  present in the sample is 0.238 g.

The reaction product is $B_4O_5(OH)4^{2^{-}} + 2H_3O+ + 3H_2O + 4H_3BO_3,$

which contains a total of 4 moles of $H_3BO_3$ and 1 mole of $B_4O_5(OH)4^{2-}$.

The molar mass of $B_4O_5(OH)4^{2-}$ is 381.35 g/mol, which means that the total mass of the reaction product can be calculated as follows:

4 mol of $H_3BO_3$ x 61.83 g/mol + 1 mol of $B_4O_5(OH)4^{2-}$ x 381.35 g/mol = 855.44 g/mol

The percentage of $B_2O_3$ in the reaction product can be calculated by dividing the mass of $B_2O_3$ by the total mass of the reaction product and multiplying by 100:

0.238 g of $B_2O_3$ / 855.44 g of reaction product x 100% = 27.81%

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