$AgNO_3 + KBr → AgBr↓ + KNO_3$
$AgNO_3 + KI → AgI↓ + KNO_3$
$n_{KBr}=0,3.0,25=0,075 (mol)$
$⇒n_{AgBr}=n_{KBr}=0,075 (mol)$
$n_{KI}=0,15.0,25=0,0375 (mol)$
$⇒n_{AgI}=n_{KI}=0,0375 (mol)$
$⇒m_{\text{ kết tủa}}=m_{AgBr}+m_{AgI}$
$=0,075.188+0,0375.235=22,9125 (g)$