STOICHIOMETRY: QUANTITATIVE INFORMATION ABOUT CHEMICAL REACTIONS

Question

How many mL of 0.427 M HBr are needed to dissolve 6.83 g of MgCO₃?

2HBr(aq) + MgCO₃(s)  MgBr₂(aq) + H₂O(l) + CO₂(g)

_______mL

Answer 1

moles MgCO₃ = 6.83 g MgCO₃ x 1 mol/84.3 g = 0.0810 moles

moles HBr required = 0.0810 mol MgCO₃ x 2 mol HBr/mol MgCO₃ = 0.162 moles

volume of 0.427 M HBr required: (x L)(0.427 mol/L) = 0.162 moles and x = 0.379 L = 379 mls HBr