Write skeleton equations for these reactions.
a. butane(l) + oxygen (g) --> carbon dioxide(g) + water(l).
b. aluminum carbonate(s)--> aluminum oxide(s) + carbon dioxide(g)
c. silver nitrate(aq) + sodium sulfide(aq) --> silver sulfide(s) + sodium nitrate(aq)
d. iron(s)+ fluorine(g) --> iron(III0 fluoride(s)
e. sulfur trioxide(g) + water(l)--> sulfuric acid(aq)
f. sodium(s) + magnesium iodide(aq)--> sodium iodide(aq) + magnesium(s)
g. vanadium(s) + oxygen(g)--> vanadium oxide(s)
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vanadium (s) + oxygen (g) arrow vanadium (v) oxide (s)
a. C4H10(l) + 13O2(g) --> 8CO2(g) + 10H2O(l)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 4 carbon atoms, 10 hydrogen atoms, and 26 oxygen atoms (13 molecules of O2).
On the right side, we have 8 carbon atoms, 20 hydrogen atoms, and 26 oxygen atoms (8CO2 and 10H2O).
To balance the carbons, we need to put a coefficient of 8 in front of CO2 on the right side.
To balance the hydrogens, we need to put a coefficient of 10 in front of H2O on the right side.
Now, the equation becomes:
C4H10(l) + 13O2(g) --> 8CO2(g) + 10H2O(l)
b. Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 2 aluminum atoms, 6 carbon atoms, and 18 oxygen atoms (3 carbonate ions, CO3).
On the right side, we have 2 aluminum atoms, 3 carbon atoms, and 9 oxygen atoms (Al2O3 and 3CO2).
The number of aluminum atoms and carbon atoms is already balanced, but the number of oxygen atoms is not.
To balance the oxygen atoms, we need to put a coefficient of 9 in front of CO2 on the right side.
Now, the equation becomes:
Al2(CO3)3(s) --> Al2O3(s) + 3CO2(g)
c. 2AgNO3(aq) + Na2S(aq) --> Ag2S(s) + 2NaNO3(aq)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 2 silver atoms, 2 nitrogen atoms, 8 oxygen atoms (2 nitrate ions, NO3), 2 sodium atoms, and 1 sulfur atom.
On the right side, we have 2 silver atoms, 1 sulfur atom, 6 oxygen atoms (Ag2S and 2 nitrate ions), and 2 sodium atoms.
The number of silver, sulfur, and sodium atoms is already balanced, but the number of nitrogen and oxygen atoms is not.
To balance the nitrogen atoms, we need to put a coefficient of 2 in front of AgNO3 on the left side.
To balance the oxygen atoms, we need to put a coefficient of 2 in front of NaNO3 on the right side.
Now, the equation becomes:
2AgNO3(aq) + Na2S(aq) --> Ag2S(s) + 2NaNO3(aq)
d. 2Fe(s) + 3F2(g) --> 2FeF3(s)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 2 iron atoms and 6 fluorine atoms (3 molecules of F2).
On the right side, we have 2 iron atoms and 6 fluorine atoms (2 FeF3).
The number of iron atoms and fluorine atoms is already balanced.
Now, the equation becomes:
2Fe(s) + 3F2(g) --> 2FeF3(s)
e. SO3(g) + H2O(l) --> H2SO4(aq)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 1 sulfur atom and 3 oxygen atoms (SO3).
On the right side, we have 1 sulfur atom, 4 oxygen atoms, and 2 hydrogen atoms (H2SO4).
The number of sulfur atoms and oxygen atoms is already balanced.
To balance the hydrogen atoms, we need to put a coefficient of 2 in front of H2O on the left side.
Now, the equation becomes:
SO3(g) + 2H2O(l) --> H2SO4(aq)
f. 2Na(s) + MgI2(aq) --> 2NaI(aq) + Mg(s)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 2 sodium atoms and 1 magnesium atom.
On the right side, we have 2 sodium atoms and 1 magnesium atom.
The number of sodium and magnesium atoms is already balanced.
Now, the equation becomes:
2Na(s) + MgI2(aq) --> 2NaI(aq) + Mg(s)
g. 4V(s) + 3O2(g) --> 2V2O3(s)
To balance this equation, we start by counting the number of atoms of each element on both sides of the equation.
On the left side, we have 4 vanadium atoms and 6 oxygen atoms (3 molecules of O2).
On the right side, we have 4 vanadium atoms and 6 oxygen atoms (2 V2O3).
The number of vanadium atoms and oxygen atoms is already balanced.
Now, the equation becomes:
4V(s) + 3O2(g) --> 2V2O3(s)