Consider the following reaction: BaCl_2(aq) + Na_2SO_4(aq) rightarrow BaSO_4(s)

Question

Consider the following reaction: BaCl_2(aq) + Na_2SO_4(aq) rightarrow BaSO_4(s) + 2 NaCl(aq) The number of moles of BaCl_2 is held constant at 6.00 times 10^-3 moles, while the number of moles of Na_2SO_4 is varied. Examine the graph below to answer the following questions. a) What is the stoichiometric point based on these data? b) Which reactant is the limiting reagent before the stoichiometric point? c) If you added 2.5 times 10^-3 mol of Na_2SO_4, calculate the theoretical yield of BaSO_4 (MW = 201.32 g/mol).

Explanation / Answer

from the graph Na2SO4 is limiting reactant before the stoichometric point which is 6 * 10-3 moles of Na2SO4

it is stoichometric point because above that point how much ever Na2SO4 we add the mass of preciptate remains constant at 1.4 g so it is independent of Na2SO4 and before that point mass of preciptate keep on changing even if we keep BaCl2 constant and change Na2SO4 hence reaction is dependent on Na2SO4 upto stoichometric poin t

Answer

a) 6 * 10-3 moles of Na2SO4

b) Na2SO4

c) upto 6 * 10-3 moles of Na2SO4 Na2SO4 is the limiting reactant so

According to reaction stoichometry 1 mole of Na2SO4 will yield 1 mole of BaSO4

so 2.5 * 10-3  moles of Na2SO4 will yield 2.5 * 10-3  moles of BaSO4

Mol.wt of BaSO4= 201.32 g/mol

Mass of BaSO4 = No. of moles * Mol. wt = 2.5 * 10-3  moles * 201.32 g/mol = 0.5033 g

Theoretical yield = 0.5033 g Answer

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