What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of a 0.150 M solution of iron(III) sulfate

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) >3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.