What mass (g) of AgBr is formed when 35.5 mL of 0.184 M AgNO3 is treated...

Question

Question

What mass (g) of AgBr is formed when 35.5 mL of 0.184 M AgNO₃ is treated with an excess of aqueous hydrobromic acid?

science chemistry

Answer 1

Answer #1

Moles of AgNO3 = molarity×volume/1000

= 0.184×35.5/1000 = 0.006532 moles

AgNO3 + HBr = AgBr + HNO3

Molar mass of AgBr = 187.77g/mol

Molar mass of AgNO3 = 169.87g/mol

1 mole of AgNO3 give 1 mole of AgBr.

0.006532 moles of AgNO3 give 0.006532 moles of AgBr

= 0.006532×187.77 g = 1.226 g

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Answer 2

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