Respuesta :
Based on the results of the titration, there are 1.995 g of antimony in the stibnite sample, and its percentage is 20.7%.
In order to solve this, we need to write the balanced ionic reaction equation used in the titration:
3Sb³⁺(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 3Sb⁵⁺(aq) + Br⁻(aq) + 3H₂O(l)
From the equation, we can tell that one mole of bromate (BrO₃⁻) reacts with 3 moles of Sb³⁺. To calculate the amount of Sb³⁺ from the sample, we must first calculate the number of moles (n) using the molarity (c = 0.1250 M) and the volume (43.70 mL = 0.0437 L) of the bromate solution.
c = n/V ⇒ n = c * V
n = 0.1250 M * 0.0437 L = 0.0054625 mol
This means that the amount of Sb³⁺ in the sample is 3 * 0.0054625 mol = 0.0163875 mol
To calculate the mass (m) of antimony in the sample, we will use the molar mass (M) of antimony (121.76 g/mol).
n = m/M ⇒ m = n*M
m = 0.0163875 mol * 121.76 g/mol = 1.995 g of antimony in the ore
Finally, we can calculate the percentage of antimony in stibnite:
%weight = 100% * m(antimony) / m(sample)
%weight = 100% * 1.995 g / 9.62 g = 20.7%
You can learn more about titrations here:
brainly.com/question/2728613
#SPJ4
