Respuesta :
Answer:
m AgBr = 1.193 g
Explanation:
- AgNO3 + HBr → AgBr + HNO3
∴ V AgNO3 = 41.8 mL = 0.0418 L
∴ [ AgNO3 ] = 0.152 mol/L
∴ Mw AgBr = 187.77 g/mol
⇒ n AgNO3 = (0.0418 L)(0.152 mol/L) = 6.3536 E-3 mol AgNO3
⇒ n AgBr = (6.3536 E-3 mol AgNO3)(mol AgBr/mol AgNO3) = 6.3536 E-3 mol AgBr
⇒ m AgBr = (6.3536 E-3 mol AgBr)(187.77 g/mol) = 1.193 g AgBr
The mass of AgBr = 1.193 g
Chemical reaction:
AgNO₃ + HBr → AgBr + HNO₃
Given:
Volume of AgNO₃ = 41.8 mL = 0.0418 L
Molarity of AgNO₃ = 0.152 mol/L
Molar mass of AgBr = 187.77 g/mol
Calculation for mass:
n AgNO₃ = (0.0418 L)(0.152 mol/L) = [tex]6.3536 * 10^{-3}[/tex] mol AgNO₃
n AgBr = ( [tex]6.3536 * 10^{-3}[/tex] mol AgNO₃)(mol AgBr/mol AgNO₃) = [tex]6.3536 * 10^{-3}[/tex] mol AgBr
m AgBr = ( [tex]6.3536 * 10^{-3}[/tex] mol AgBr)(187.77 g/mol) = 1.193 g AgBr
Thus, the mass of AgBr is 1.193 g.
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