Answer to Question #86046 in General Chemistry for thomas

Question

How many mL of 0.683 M HI are needed to dissolve 7.09 g of CaCO3?

2HI(aq) + CaCO3(s) CaI2(aq) + H2O(l) + CO2(g)

_________ mL

Answer 1

2HI(aq)+CaCO₃(s)->CaCl₂(aq) +H₂O(l)+CO₂(g)

n=m/M

n(CaCO₃)= 7.09 g/100.09 g/mol = 0.0708 mol

According to equation mole ratio n(HI):n(CaCO₃)= 2:1, then n(HI)= 2*n(CaCO₃)= 0.0708*2= 0.142 mol.

c=n/V, then V= n/c= 0.142 mol/0.683 mol/L= 0.207L =207 ml

Answer: 207 ml