To determine how much phosphine ([tex]\(PH_3\)[/tex]) can be prepared when 1 mole of calcium phosphide ([tex]\(Ca_3P_2\)[/tex]) reacts with an excess of water ([tex]\(H_2O\)[/tex]), we need to use stoichiometry based on the given chemical equation:
[tex]\[
Ca_3P_2 + 6 H_2O \rightarrow 3 Ca(OH)_2 + 2 PH_3
\][/tex]
Here are the steps to solve the problem:
1. Write down the balanced chemical equation:
[tex]\[
Ca_3P_2 + 6 H_2O \rightarrow 3 Ca(OH)_2 + 2 PH_3
\][/tex]
2. Identify the mole ratio between reactants and products:
From the balanced equation, we can see the mole ratio between calcium phosphide ([tex]\(Ca_3P_2\)[/tex]) and phosphine ([tex]\(PH_3\)[/tex]). Specifically, 1 mole of [tex]\(Ca_3P_2\)[/tex] produces 2 moles of [tex]\(PH_3\)[/tex].
3. Determine the amount of phosphine produced:
Since we are starting with 1 mole of calcium phosphide and there is excess [tex]\(H_2O\)[/tex], we use the mole ratio from the balanced equation:
[tex]\[
1 \text{ mole of } Ca_3P_2 \rightarrow 2 \text{ moles of } PH_3
\][/tex]
Therefore, if we start with 1 mole of [tex]\(Ca_3P_2\)[/tex], it will produce:
[tex]\[
1 \text{ mole of } Ca_3P_2 \times \frac{2 \text{ moles of } PH_3}{1 \text{ mole of } Ca_3P_2} = 2 \text{ moles of } PH_3
\][/tex]
So, the amount of phosphine ([tex]\(PH_3\)[/tex]) that can be prepared is 2 moles.
