5g of calcium carbonate and 50cm^3 of 1.0M hydrochloric acid react together.

i) write a blanced equation for the reaction.
ii) What is the amount(moles) of each is added?
iii) which is the limiting reagent?
iv) calculate the maximum volume of carbon dioxide gas produced.

i) CaCO3 + 2 HCl ------> CaCl2 + CO2 + H2O
ii) moles HCl: 0.050 L x 1.0M = ?
moles CaCO3: 5 g CaCO3/ m.w. CaCO3= ??
iii) from bal. rxn., 1 mole of CaCO3 reacts with 2 moles of HCl (1:2)
from moles of reactants and knowing you need twice as many moles of HCl as CaCO3, you determine which is limiting reactant
iv) if limiting reactant is HCl: moles of HCl x 1/2 = moles CO2
if limiting reactant is CaCO3: moles of CaCO3 = moles of CO2
assuming STP conditions, moles CO2 x 22.4 L/mole = answer

You should be able to do the equation pretty simply for yourself, but for the sake of the answer here it is.

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

Moles is mass / Molar Mass for the solid, and Conc x Vol for the liquid

n = 5g / 100.09 g/mol = 0.04996 mol

n = 1.0 mol/L x 50 mL = 50 millimoles = 0.050 mol

Limiting reagent: CaCO3. Notice that the mol ratio between the reactants is 1 to 1. So if you have 0.04996 mol of CaCO3 you need 0.04996 mol of HCl to completely use it up. You have 0.050 mol, therefore you have enough HCl plus a little bit more.

Vol of CO2. Well, you don't say what conditions, so I'll assume STP. The mol ration between CaCO3 and CO2 is also 1 to 1 so you will produce 0.04996 mol of CO2 . At STP that amounts to 0.0996 mol x 22.4 L/mol = 1.12 L