Question

How many mL of 0.523 M HNO3 are needed to dissolve 8.90 g of MgCO3?

2HNO3(aq) + MgCO3(s) Mg(NO3)2(aq) + H2O(l) + CO2(g)

Answer 1

Molar mass of MgCO3,

MM = 1*MM(Mg) + 1*MM(C) + 3*MM(O)

= 1*24.31 + 1*12.01 + 3*16.0

= 84.32 g/mol

mass(MgCO3)= 8.90 g

number of mol of MgCO3,

n = mass of MgCO3/molar mass of MgCO3

=(8.9 g)/(84.32 g/mol)

= 0.1056 mol

From the reaction

2HNO3(aq) + MgCO3(s) Mg(NO3)2(aq) + H2O(l) + CO2(g)

moles of HNO3 reacting = 2*number of mol of MgCO3

= 2*0.1056 mol

= 0.2112 mol

Now we should use:

moles of HNO3 = Molarity * volume

0.2112 mol = 0.523 M * V

V = 0.404 L

V = 404 mL

Anwer: 404 mL