Question

171.00 mL of 0.400 M AgNO₃(aq) is mixed with 113.00 mL of 0.100 M Sr(OH)₂(aq). Assuming 100% yield, what mass (in g) of precipitate will form?

Answer 1

To calculate the mass of the precipitate that will form when 200 mL of 0.5M NaCl and 800 mL of 0.10M AgNO3 are mixed together, we need to calculate the moles of each reactant using their concentrations and volumes. By using the balanced equation, we can relate the moles of reactants to the moles of the precipitate, AgCl. Finally, we can calculate the mass of AgCl using its molar mass.

Step-by-step explanation:

To determine the mass of the precipitate that will form when 200 mL of 0.5M NaCl and 800 mL of 0.10M AgNO3 are mixed, we need to first calculate the moles of each reactant. Moles can be calculated using the formula: moles = concentration (M) x volume (L).

For NaCl:

Moles of NaCl = 0.5M x 0.2L = 0.1 moles

For AgNO3:

Moles of AgNO3 = 0.1M x 0.8L = 0.08 moles

Since the balanced equation states that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl, we can determine the mass of AgCl using its molar mass.

Mass of AgCl = Moles of AgCl x Molar mass of AgCl

From the equation, we know that 1 mole of AgCl has a molar mass of 143.32 g/mol. Therefore:

Mass of AgCl = 0.08 moles x 143.32 g/mol = 11.46 grams