Asked by Freddy
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)
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)
Answers
Answered by
DrBob222
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 = ?
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 = ?
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