Question

How many grams of NaOH is required to neutralize 54.75 grams of HCL?

Answer 1

Answer:

Explanation:

NaOH +HCL  = NaCL+H₂O

THAT IS 1 MOLE OF Na OH IS REQUIRED FOR THE NEUTRALIAZATION OF 1 MOLE OF HCL.

1 MOLE OF NaOH WEIGHS = 23+16+1

                                               = 40 G

1 MOLE OF HCL WEIGHS = 35.5+1

                                          = 36.5 G

THAT IS ;

40 G NaOH NEUTRALIZES 36.5 G HCL

THEREFORE 54.75 GRAMS OF HCL IS NEUTRALIZED BY =54.75*40/36.5

     = 60G NaOH

1 MOLE NaON = 40 G

THERFORE 60 G NaOH = 60/40

                                         = 1.5 MOLES OF NaOH IS REQUIRED TO NEUTRALIZE 54.75 G HCL.

HOPE THIS HELPS.