Complete and balance the following equations.Identify each type of the reactions. (a) Cr + O2 >Cr2O3: (b)Na2CO3 + Mg3(PO4)2: (c)H3BCO+ H2O> B(OH)3 + CO + H2:
(a) The given equation is: Cr + O2 → Cr2O3
This is a combustion reaction since oxygen (O2) reacts with chromium (Cr) to form chromium(III) oxide (Cr2O3).
To balance this equation, follow the step-by-step process:
Step 1: Count the number of atoms in each element on both sides of the equation.
Left side: Cr: 1 atom, O: 2 atoms
Right side: Cr: 2 atoms, O: 3 atoms
Step 2: Start with the element that appears in the fewest molecules/compounds on either side. In this case, it's Cr, which appears as a single atom on the left side and in a compound with two atoms on the right side.
Step 3: Balance the Cr atoms by placing a coefficient of 2 in front of Cr on the left side. Now the equation becomes:
2Cr + O2 → Cr2O3
Step 4: Revaluate the number of atoms on both sides:
Left side: Cr: 2 atoms, O: 2 atoms
Right side: Cr: 2 atoms, O: 3 atoms
Step 5: The oxygen atoms are unbalanced. To balance them, place a coefficient of 3/2 (1.5) in front of O2:
2Cr + (3/2)O2 → Cr2O3
Since we can't have a fractional coefficient, multiply the entire equation by 2 to eliminate the fraction:
4Cr + 3O2 → 2Cr2O3
The balanced equation is: 4Cr + 3O2 → 2Cr2O3
(b) The given equation is: Na2CO3 + Mg3(PO4)2
This is a double displacement reaction, also known as a precipitation reaction, where sodium carbonate (Na2CO3) reacts with magnesium phosphate (Mg3(PO4)2).
To balance this equation, ensure that the same number of each type of atom is present on both sides. However, note that this equation is not complete; there is no product.
Without a product, it's not possible to balance the equation properly or determine the reaction type accurately.
Please provide the missing product, and I'll be happy to assist you in balancing the equation and identifying the reaction type.
(c) The given equation is: H3BCO + H2O → B(OH)3 + CO + H2
This is a decomposition reaction, where boric acid carbonate (H3BCO) reacts with water (H2O) to produce boron hydroxide (B(OH)3), carbon monoxide (CO), and hydrogen gas (H2).
To balance this equation, follow the step-by-step process:
Step 1: Count the number of atoms in each element on both sides of the equation.
Left side: H: 5 atoms, B: 1 atom, C: 1 atom, O: 3 atoms
Right side: B: 1 atom, H: 5 atoms, O: 4 atoms, C: 1 atom
Step 2: Start with the most complex compound that appears on one side only, which in this case is H3BCO on the left side.
Step 3: Balance the atoms in H3BCO by placing a coefficient of 1 in front of H3BCO on the left side. Now the equation becomes:
H3BCO + H2O → B(OH)3 + CO + H2
Step 4: Reevaluate the number of atoms on both sides:
Left side: H: 5 atoms, B: 1 atom, C: 1 atom, O: 3 atoms
Right side: B: 1 atom, H: 7 atoms, O: 5 atoms, C: 1 atom
Step 5: The hydrogen atoms are unbalanced. To balance them, place a coefficient of 2 in front of H2O:
H3BCO + 2H2O → B(OH)3 + CO + H2
Step 6: Reevaluate the number of atoms on both sides:
Left side: H: 7 atoms, B: 1 atom, C: 1 atom, O: 5 atoms
Right side: B: 1 atom, H: 7 atoms, O: 5 atoms, C: 1 atom
All the atoms are balanced, and the balanced equation is:
H3BCO + 2H2O → B(OH)3 + CO + H2