Final answer:
The stoichiometry of the reaction between metal bromide, MBrn, and silver nitrate, AgNO3, shows that the mole ratio is 1:2, leading to the formula of the metal bromide being MBr2.
Step-by-step explanation:
To determine the formula of the metal bromide, MBrn, we need to use the stoichiometry of the reaction with silver nitrate, AgNO3. Silver nitrate reacts with bromide ions to form silver bromide, AgBr, a precipitate. The balanced chemical equation for this reaction is:
AgNO3 + MBrn → AgBr + M(NO3)n
From the question data, we have 160 cm³ of a 0.100 mol dm³ AgNO3 solution, which means there is 0.016 mol of AgNO3 present (since 160 cm³ is 0.16 dm³ and 0.16 dm³ × 0.100 mol dm³ = 0.016 mol).
The 20.0 cm³ of the metal bromide solution has a concentration of 0.400 mol dm³, meaning there are 0.008 moles of MBrn present (since 20 cm³ is 0.020 dm³ and 0.020 dm³ × 0.400 mol dm³ = 0.008 mol).
To react exactly with the metal bromide, the mole ratio should match the coefficients in the balanced chemical equation. Since all of the AgNO3 reacts, and the amount present is twice the amount of MBrn, it suggests that n in MBrn must be 2. Therefore, for every mole of MBr2, two moles of AgNO3 are required. This leads to the formula of the metal bromide being MBr2. Without the exact mass of the metal, we cannot determine the specific metal.