Calculate the amount of substance (mol) and the mass of each product formed during the following transformations
Calculate the amount of substance (mol) and the mass of each product formed during the following transformations: sulfur → sulfur (IV) oxide → sulfurous acid → -barium sulfite, if 16 g of sulfur were taken.
Given:
m (S) = 16 g
Find:
n (SO2) -?
m (SO2) -?
n (H2SO3) -?
m (H2SO3) -?
n (BaSO3) -?
m (BaSO3) -?
n (H2O) -?
m (H2O) -?
Solution:
1) S + O2 => SO2;
SO2 + H2O => H2SO3;
H2SO3 + Ba (OH) 2 => BaSO3 + 2H2O;
2) n (S) = m (S) / Mr (S) = 16/32 = 0.5 mol;
3) n (SO2) = n (S) = 0.5 mol;
4) m (SO2) = n (SO2) * Mr (SO2) = 0.5 * 64 = 32 g;
3) n (H2SO3) = n (SO2) = 0.5 mol;
4) m (H2SO3) = n (H2SO3) * Mr (H2SO3) = 0.5 * 82 = 41 g;
3) n (BaSO3) = n (H2SO3) = 0.5 mol;
4) m (BaSO3) = n (BaSO3) * Mr (BaSO3) = 0.5 * 217 = 108.5 g;
3) n (H2O) = n (H2SO3) * 2 = 0.5 * 2 = 1 mol;
4) m (H2O) = n (H2O) * Mr (H2O) = 1 * 18 = 18 g.
Answer: The amount of SO2 is 0.5 mol; H2SO3 – 0.5 mol; BaSO3 – 0.5 mol; H2O – 1 mol; the mass of SO2 is 32 g; H2SO3 – 41 g; BaSO3 – 108.5 g; H2O – 18 g.