88 g of the strontium reacts with 160 g bromine to produce how many strontium bromide

Mass of strontium = 88 g

Atomic mass of Sr = 87.62 u

Moles = mass /atomic mass

Moles of Sr =  88 / 87.62 = 1.004

Mass of bromine = 160 g

Atomic mass of Br = 79.904 u

Molar mass of Br₂ = 79.904 x 2 = 159.808 g/mol

Moles = mass /molar mass

Moles of Br₂ =  160 / 159.808 = 1.001

The balanced chemical equation of the reaction between strontium and bromine:

Sr(s) + Br₂(l) = SrBr₂

The molar ratio between Br₂  and SrBr₂ is 1:1

So the moles of SrBr₂ produced from 1.001 moles of Br₂ is 1.001.

Moles = mass /molar mass

Mass = moles x molar mass

Molar mass of SrBr₂ = 247.4280 g/mol

Mass = 1.001 mol x 247.4280 g/mol

Mass = 247.428 g

The mass of strontium bromide produced is 247.428 g.

the ratio of the conjugate base to the acid required to attain the desired ph.