ZnCO3 (g) + 2HNO3 (aq) Zn (NO3) 2 + CO2 (g) + H2O (l)
millimoles HNO3 = mL x M = 50 x 4 = 200
That 200 mmols HNO3 will use 1/2 x 200 = 100 mmols Zn or 0.1 mols Zn.
How much Zn did you have? That's mols Zn = g/atomic mass = 17.0/65.4 0.26. You used 0.1. Remaining Zn = 0.26- 0.1 = 0.16 mols. grams Zn = 0.16 mols x 65.4 = ?
4. Calculate the mass of zinc carbonate that would remain if 17.0g of zinc carbonate was
reacted with 50cm3 of 4M nitric acid. The equation of the reaction is:
ZnCO3 (g) + 2HNO3 (aq) Zn (NO3) 2 + CO2 (g) + H2O (l)
(Zn=65.4, C=12.0, O=16.0)
A certain carbonate XCO3 , reacts with dilute hydrochloric acid according to the
equation given below:
XCO3(s) +2HCl (aq) XCl2 (aq) + CO2 (g) + H2O (l)
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