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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 reaction1ca(oh)₂ +2h₃po₃ →3ca₃(po₃)₂(s) +4h₂othis is an 5 reaction and also a 6 reaction (use the order given in the answers).is this a redox reaction?(yes/no)7, because neither ca⁺², oh⁻¹, h⁺¹, nor po₃⁻³change oxidation state.will the reaction happen as it is written? (yes/no)8, because acids and bases are reactive, the products are stable, and 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. 8i. 9 j. 10 k. 12 l. 14 m. 16 n. 18 o. 20 p. 21q. 24 r. 26 s. 28 t. synthesis u. decompositionv. single displacement w. double displacement x. neutralizationy. combustion z. yes aa. no bb. reactive cc. stabledd. oxidized ee. reduced

Explanation:

Step1: Balance the chemical equation

For \(Ca\): There are \(3\) \(Ca\) atoms in \(Ca_3(PO_3)_2\). So we need \(3\) \(Ca(OH)_2\).
For \(P\): There are \(2\) \(P\) atoms in \(Ca_3(PO_3)_2\). So we need \(2\) \(H_3PO_3\).
For \(H\) and \(O\): After putting \(3\) \(Ca(OH)_2\) and \(2\) \(H_3PO_3\), we have \(3\times2 + 2\times3=12\) \(H\) atoms and \(3\times2+2\times3 = 12\) \(O\) atoms on the left. On the right, in \(Ca_3(PO_3)_2\) we have \(2\times3 = 6\) \(O\) atoms. So for \(H_2O\), we have \(6\) \(H_2O\) (since \(6\times2 = 12\) \(H\) atoms and \(6\times1=6\) \(O\) atoms). The balanced equation is \(3Ca(OH)_2+2H_3PO_3
ightarrow Ca_3(PO_3)_2 + 6H_2O\).

Step2: Identify the reaction type

This is a double - displacement reaction (\(W\)) because the \(Ca^{2 +}\) and \(H^{+}\) ions exchange partners. It is also a neutralization reaction (\(X\)) as an acid (\(H_3PO_3\)) and a base (\(Ca(OH)_2\)) react to form a salt (\(Ca_3(PO_3)_2\)) and water.

Step3: Determine if it is a redox reaction

It is not a redox reaction (\(AA\)) because the oxidation states of \(Ca\) (\(+ 2\)), \(O\) (\(-2\) in \(OH^-\) and \(PO_3^{3 -}\)), \(H\) (\(+1\)) and \(P\) (\(+3\) in \(PO_3^{3 -}\)) do not change.

Step4: Determine if the reaction occurs as written

The reaction will happen as written (\(Z\)) because acids and bases are reactive, the products are stable, and one product (\(Ca_3(PO_3)_2\)) is a solid (driving the reaction forward according to the solubility rules).

Answer:

  1. \(C\) (for \(3Ca(OH)_2\))
  2. \(B\) (for \(2H_3PO_3\))
  3. \(A\) (for \(Ca_3(PO_3)_2\))
  4. \(I\) (for \(6H_2O\))
  5. \(X\) (neutralization)
  6. \(W\) (double displacement)
  7. \(AA\) (no)
  8. \(Z\) (yes)