The mass fractions of sodium, oxygen, chlorine and hydrogen in substance X are 14.0, 58.3, 21.6 and 6.1%, respectively.

| February 13, 2021 | education

The mass fractions of sodium, oxygen, chlorine and hydrogen in substance X are 14.0, 58.3, 21.6 and 6.1%, respectively. How many grams of bromine are formed when 10.0 g of X is treated with an excess of hydrobromic acid? Take the relative atomic mass of chlorine to be 35.5.

Given: substance X ω (Na) = 14% ω (O) = 58.3% ω (Cl) = 21.6% ω (H) = 6.1% m (in-va X) = 10 g Find: m (Br2) -?
Decision:
1) Find the mass of each element in substance X: m (Na) = ω (Na) * m (in-va X) / 100% = 14% * 10/100% = 1.4 g; m (O) = ω (O) * m (in-va X) / 100% = 58.3% * 10/100% = 5.83 g; m (Cl) = ω (Cl) * m (in-va X) / 100% = 21.6% * 10/100% = 2.16 g; m (H) = ω (H) * m (in-va X) / 100% = 6.1% * 10/100% = 0.61 g; 2) Find the amount of substance of each element in substance X: n (Na) = m (Na) / Аr (Na) = 1.4 / 23 = 0.06 mol; n (O) = m (O) / Ar (O) = 5.83 / 16 = 0.36 mol; n (Cl) = m (Cl) / Ar (Cl) = 21.6 / 35.5 = 0.06 mol; n (H) = m (H) / Ar (H) = 0.61 / 1 = 0.61 mol; 3) Find the formula of substance X: The ratio of the amount of substance between the elements in substance X will correspond to their index in substance X.Then n (Na): n (O): n (Cl): n (H) = 0.06: 0, 36: 0.06: 0.61 = 1: 6: 1: 10; NaO6ClH10, or NaClO * 5H2O; 4) Write the reaction equation: NaClO * 5H2O + 2HBr => Br2 + NaCl + 6H2O 5) Find the amount of the substance NaClO * 5H2O (by definition): n (NaClO * 5H2O) = n (Na) = 0.06 mol; 6) Find the amount of substance Br2 (taking into account the reaction equation): n (Br2) = n (NaClO * 5H2O) = 0.06 mol; 7) Find the mass of Br2: m (Br2) = n (Br2) * Mr (Br2) = 0.06 * 160 = 9.6 g.
Answer: The mass of Br2 is 9.6 g.



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