QUESTION IMAGE
Question
balance the reaction, do not leave any fractions, dont leave anything blank, then answer the questions about the reaction
_1_ba(cn)₂ +_2_na₃po₄ →_3_nacn +_4_ba₃(po₄)₂(s)
this is an _5_ reaction. is this a redox reaction? (yes/no)_6_, because neither ba⁺², cn⁻¹, na⁺¹, nor po₄⁻³ change oxidation state.
will the reaction happen as it is written? (yes/no)_7_, because one product is a solid. the reverse reaction would not happen.
a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8
i. 9 j. 10 k. 12 l. 14 m. 16 n. 18 o. 20 p. 21
q. 24 r. 26 s. 28 t. synthesis u. decomposition
v. single displacement w. double displacement x. neutralization
y. combustion z. yes aa. no bb. reactive cc. stable
dd. oxidized ee. reduced
Step1: Balance Ba atoms
On the right - hand side, there are 3 Ba atoms in \(Ba_3(PO_4)_2\). So, we put a coefficient of 3 in front of \(Ba(CN)_2\).
Step2: Balance \(PO_4\) groups
On the right - hand side, there are 2 \(PO_4\) groups in \(Ba_3(PO_4)_2\). So, we put a coefficient of 2 in front of \(Na_3PO_4\).
Step3: Balance Na atoms
On the left - hand side, there are \(2\times3 = 6\) Na atoms in \(2Na_3PO_4\). So, we put a coefficient of 6 in front of \(NaCN\).
For the reaction type:
In a double - displacement reaction, the general form is \(AB + CD
ightarrow AD+CB\). Here, \(Ba(CN)_2\) (\(AB\) where \(A = Ba^{2 +}\) and \(B=(CN)^{-}\)) and \(Na_3PO_4\) (\(CD\) where \(C = Na^{+}\) and \(D = PO_4^{3-}\)) react to form \(NaCN\) (\(AD\)) and \(Ba_3(PO_4)_2\) (\(CB\)). So, it is a double - displacement reaction (W).
For the redox reaction:
Since there is no change in the oxidation states of \(Ba^{2+}\), \(CN^{-}\), \(Na^{+}\), and \(PO_4^{3 -}\) (as given in the problem), it is not a redox reaction (AA).
For whether the reaction will happen:
Because one of the products (\(Ba_3(PO_4)_2\)) is a solid (precipitate), the reaction will happen as written (Z).
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- C (3)
- B (2)
- F (6)
- A (1)
- W (double displacement)
- AA (no)
- Z (yes)