How many grams Ir2O3 (MW = 432 g/mol) are produced from 199 grams IrCl3 (MW = 299 g/mol)?

Question

Given the following reaction: IrCl3 + NaOH → Ir2O3 + HCl + NaCl

How many grams Ir₂O₃ (MW = 432 g/mol) are produced from 199 grams IrCl₃ (MW = 299 g/mol)?

Answer 1

Hello,

First you need to make sure that the equation in balanced in this problem and any problems similar to this. The equation balanced is : 2 IrCl₃ + 3 NaOH → Ir₂O₃ + 3 HCl + 3 NaCl

Now using the mole ratio information, it is telling you that for every 2 moles of IrCl₃ used, one mole of Ir₂O₃ will be produced.

The amount of reactant is given in grams and you should know how to get moles of a reactant from grams (÷ by the molar mass).

The steps are : g reactant → mol reactant →mol product → g product

with each arrow representing a step using a conversion relationship.

199 g IrCl₃ (1 mol IrCl₃/ 299 g IrCl₃) ( 1 mol Ir₂O₃/ 2 mol IrCl₃) ( 432 g Ir₂O₃ / 1 mol Ir₂O₃) = 144 g

You can use the above information to help solving any problem similar to this one.

All the best!