0.100 M AlBr3 37.5 mL AgNO3 AgBr mass

0.100 M 브롬화 알루미늄 용액 37.5 mL가

질산은 용액과 다음과 같이 반응한다면,

AgBr(187.77 g/mol)의 생성량은 얼마인가?

AlBr3(aq) + 3AgNO3(aq) → 3AgBr(s) + Al(NO3)3(aq)

Given that 37.5 mL of 0.100 M aluminum bromide solution reacts with a silver nitrate solution, what is the mass of AgBr (187.77 g/mol) produced?

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(0.100 mol/L) (37.5/1000 L)

= 0.00375 mol AlBr3

AlBr3(aq) + 3AgNO3(aq) → 3AgBr(s) + Al(NO3)3(aq)

AlBr3 : AgBr = 1 : 3 = 0.00375 mol : ? mol

? = 3 × 0.00375 = 0.01125 mol AgBr

AgBr의 몰질량 = 187.77 g/mol 이므로,

0.01125 mol × (187.77 g/mol)

= 2.1124 g AgBr

답: 2.11 g

[키워드] AlBr3 + AgNO3 반응 기준, AgBr 침전 기준