Question

a mixture consisting of only chromium(iii) bromide (crbr3) and aluminum bromide (albr3) weighs 0.9402 g. when the mixture is dissolved in water and an excess of silver nitrate is added, all the bromide ions associated with the original mixture are precipitated as insoluble silver bromide (agbr). the mass of the silver bromide is found to be 1.9010 g. calculate the mass percentages of chromium(iii) bromide and aluminum bromide in the original mixture.

Answer 1

The mass percentages of CrBr3 and AlBr3 in the mixture can be determined by calculating the moles of AgBr, which gives the moles of bromide ions, and then using the stoichiometry of the two compounds to find the moles and subsequently the masses and mass percentages of CrBr3 and AlBr3 in the original mixture.

Step-by-step explanation:

To calculate the mass percentages of chromium(iii) bromide (CrBr3) and aluminum bromide (AlBr3) in the original mixture, we first need to determine the number of moles of silver bromide (AgBr) that precipitated. Given that the mass of AgBr is 1.9010 g, we can calculate the moles of AgBr using its molar mass (187.77 g/mol).

Next, because each mole of AgBr corresponds to one mole of Br−, we divide the number of moles of AgBr by the stoichiometry of CrBr3 and AlBr3 to find the moles of each compound in the mixture. Since CrBr3 has three moles of Br− per mole of compound and AlBr3 also has three moles of Br−, we can deduce that the molar amounts of CrBr3 and AlBr3 would be equal to the moles of Br− divided by three.

Finally, we can calculate the respective masses of CrBr3 and AlBr3 by multiplying the moles of each compound by their molar masses, and then we find the mass percentages by dividing these masses by the total mass of the mixture (0.9402 g) and multiplying by 100.