What mass of solid agbr is produced when 100.0 ml of 0.150 m agno3 is added to 20.0 ml of 1.00 m nabr?

First check for the limiting reagent. You need the balanced reaction and you need to add up the molar masses. AgNO3 + NaBr > AgBr + NaNO3 You convert the masses to moles and divide each by the corresponding coefficient in the balanced reaction, however all of the coefficients are 1. (.150 mol/L AgNO3)(.1000 L) = .0150 mol AgNO3 (1.00 mol/L NaBr)(.0200 L) = .0200 mol NaBr The AgNO3 is the limiting reagent. The balanced reaction tells you that the number of moles of AgBr produced is the same as the number of moles of AgNO3 that react, which is .0150 moles. Change back to grams by multiplying by the molar mass (.0150 mol AgBr)(188 g/mol) = 2.82 g AgBr

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