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balance the reaction, do not leave any fractions, dont leave anything b…

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

Explanation:

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 \(Ba(CN)_2\).

$$3Ba(CN)_2+\_Na_3PO_4 ightarrow\_NaCN + Ba_3(PO_4)_2$$

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 \(Na_3PO_4\).

$$3Ba(CN)_2 + 2Na_3PO_4 ightarrow\_NaCN+Ba_3(PO_4)_2$$

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 \(NaCN\).

$$3Ba(CN)_2+2Na_3PO_4 ightarrow6NaCN + Ba_3(PO_4)_2$$

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).

Answer:

  1. C (3)
  2. B (2)
  3. F (6)
  4. A (1)
  5. W (double displacement)
  6. AA (no)
  7. Z (yes)