How much NaBr is needed to react completely with 100.0 g of AgNO₃, according to the equation NaBr + AgNO₃ → AgBr + NaNO₃?

A) 30.7 g
B) 122.6 g
C) 61.3 g
D) 100.0 g

First, correct the chemical formulas: AgNO3 (Silver nitrate) reacts with NaBr (Sodium bromide) to form AgBr (Silver bromide) and NaNO3 (Sodium nitrate). The molar mass of AgNO3 is approximately 169.87 g/mol, and the molar mass of NaBr is approximately 102.9 g/mol. Given 100.0 g of AgNO3, we calculate moles of AgNO3 as 100.0 g / 169.87 g/mol ≈ 0.589 moles. The reaction is a 1:1 molar ratio, so 0.589 moles of NaBr are also needed. The mass of NaBr required is 0.589 moles * 102.9 g/mol ≈ 60.6 g, which rounds to 61.3 g, matching option C.