Practice Homework 40: Predicting Products – Metal Hydrogen Carbonate Decomposition
General Information

See Notes-47 for help. The textbook does not cover this material.

This assignment is for practice only. It is not due.

Remember that it helps to name the product before writing its chemical formula.

Please balance the equations.
Definitions/Helpful Info

Assume incomplete metal hydrogen carbonate decomposition.

Metal hydrogen carbonate(s) –(heat) metal carbonate(s) + CO2(g) + H2O(g)

Assume that transition metals retain the same charge on both sides of the word equation.
Part 1: Complete the word equations, given the reactants. (Do not write the skeleton equations).
∆
1)
sodium bicarbonate(s)
→
2)
magnesium hydrogen carbonate(s)
→
3)
aluminum hydrogen carbonate(s)
→
4)
tin(IV) bicarbonate(s)
→
5)
niobium(V) hydrogen carbonate(s)
→
∆
∆
∆
∆
Page 1 of 3
Part 2: Complete the chemical equations below. Balance the equations
Remember:
Step 1) Name the products in your head, or write it down. (Remember: just cross out the “bi” in “bicarbonate” to get the name of the main
product).
Step 2) Write the chemical formulas from the names.
Step 3) Balance the equation: Balance the metal first, then balance hydrogen, carbon dioxide gets the same coefficient as water.
∆
6)
Pb(HCO3)4(s)
→
7)
Ca(HCO3)2(s)
→
8)
V(HCO3)5(s)
→
9)
KHCO3(s)
→
10)
V(HCO3)3(s)
→
∆
∆
∆
∆
* Answers on the next page.
Page 2 of 3
Answers
∆
sodium carbonate(s) + carbon dioxide(g) + water(g)
∆
magnesium carbonate(s) + carbon dioxide(g) + water(g)
∆
aluminum carbonate(s) + carbon dioxide(g) + water(g)
∆
tin(IV) carbonate(s) + carbon dioxide(g) + water(g)
∆
niobium(V) carbonate(s) + carbon dioxide(g) + water(g)
1)
sodium bicarbonate(s)
→
2)
magnesium hydrogen carbonate(s)
→
3)
aluminum hydrogen carbonate(s)
→
4)
tin(IV) bicarbonate(s)
→
5)
niobium(V) hydrogen carbonate(s)
→
Part 2: Complete the chemical equations below. Balance the equations
6)
(1)
Pb(HCO3)4(s)
∆
→
(1) Pb(CO3)2(s) + 2 CO2(g) + 2 H2O(l)
* Remember: You don’t have to write “1” for coefficients.
∆
(1) CaCO3(s) + (1) CO2(g) + (1) H2O(l)
∆
(1) V2(CO3)5(s) + 5 CO2(g) + 5 H2O(l)
∆
(1) K2CO3(s) + (1) CO2(g) + (1) H2O(l)
7)
(1)
Ca(HCO3)2(s)
→
8)
2
V(HCO3)5(s)
→
9)
2
KHCO3(s)
→
10)
2
V(HCO3)3(s)
→ (1) V2(CO3)3(s) + 3 CO2(g) + 3 H2O(l)
∆
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