In the reaction given in problem 5, 80.0 mL of 2.0 M HCl would react with how many grams of aluminum?

Question

dominantvampire · Answer

45.0 g of Ca(NO3)2 was used to create a 1.3 M solution. What is the volume of the solution?

anonymous · Answer

use the simple molarity formula. Concentration = Moles / volume in litre
since 45 g of Ca(NO3)2 was used to prepare 1.3 M solution, 13 M = 45 g/ Mol wt of ca(NO3)2 * Volume

Solvin for volume will get you the answer.

dominantvampire · Answer

alright and 1st one? please

dominantvampire · Answer

@nanosaravana

anonymous · Answer

Every one mole o aluminium consumes 3 moles of chloride to form AlCl3. Calculate the number of moles of Cl present in the HCL colution. This will help you to figure out the required amount of Aluminium which will be one third of the Cl ions.