Question

cate grame de AgBr se formeaza prin reactia a 45 mL solutie CaBr2 0,842 M cu un exces de AgNO3?

Answer 1

Cm=n(niu)/Vs ==> n=Cm*Vs=0.842*0.045=0.03789 moli CaBr2

1mol................................................2mol
CaBr2 + 2AgNO3 --> Ca(NO3)2 + 2AgBr
0.03789 mol...................................x=0.07578

MAgBr=AAg+ABr=108+80=188g/mol

n=m/M==>m=n*M=0.07578*188=14.24664g AgBr