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------------------------------------------------ 11) 15.0 g of reactant A (FW = 102 g/mol) is mixed with 50.0 mL of 3.0 M reactant B (FW = 52.3 g/mol) and reacts according to the balanced reaction below. Determine how many grams of C (FW = 98.5 g/mol) are produced.

(Note: The balanced reaction needs to be provided for a complete solution.)

Answer :

Final answer:

Mass of C = moles of C x molar mass of C = 0.15 moles x 98.5 g/mol = 14.775 grams

Explanation:

Given:

  • Mass of reactant A = 15.0 g
  • Molar mass of reactant A (FW) = 102 g/mol
  • Volume of reactant B = 50.0 mL
  • Concentration of reactant B = 3.0 M
  • Molar mass of reactant B (FW) = 52.3 g/mol
  • Molar mass of product C (FW) = 98.5 g/mol

First, calculate the moles of reactant A:

Moles of A = mass of A / molar mass of A = 15.0 g / 102 g/mol = 0.147 moles

Then, calculate the moles of reactant B:

Moles of B = concentration of B x volume of B = 3.0 M x 0.050 L = 0.15 moles

Since reactant B has a higher number of moles (0.15 moles) than reactant A (0.147 moles), B is the limiting reactant. Therefore, the moles of C produced will be equal to the moles of B.

Finally, convert the moles of C to grams:

Mass of C = moles of C x molar mass of C = 0.15 moles x 98.5 g/mol = 14.775 grams

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