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