Calcium phosphine (Ca3P2) and water react to form calcium hydroxide and phosphin

Question

Calcium phosphine (Ca3P2) and water react to form calcium hydroxide and phosphine (PH3). In a particular experiment, 225.0g Ca3P2 and 125.0g water are combined.

Ca3P2(s) + H2O = Ca(OH)2(aq) + PH3(g)

How much PH3 can be produced?

How much Ca(OH)2 can be produced?

How much of the excess reactant remains when the reaction is complete?

I need step by step problem solving. I dont know what I am doing wrong, but for the first question I got 83.965g of PH3, according to the book answer this is wrong? :(

Explanation / Answer

The balanced reaction for this will be

Ca3P2 + 6H2O ---> 3Ca(OH)2 + 2PH3

1 mole of Ca3P2 reacts with 6 moles of water to produce 3 moles of calcium hydroxide and 2 moles of PH3

225.0 g of Ca3P2 is 225.0 / 182.18 = 1.24 moles

125.0 g of water = 125.0 / 18 = 6.94 moles

So we see water is the limiting reagent here.

6 mole of water will produce 2 moles of PH3

so 6.94 moles of water will produce ( 6.94 x 2) / 6 = 2.31 moles of PH3

molar mass of PH3 is 34 g / mol

so 2.31 moles = 2.31x 34 = 78.65 g

So 78.65 g of PH3 will be produced.

likewise 6 moles of water gives 3 moles of Ca(OH)2

so 6.94 moles of water will give ( 6.94 x 3 ) / 6 = 3.47 moles of calcioum hydroxide

molar mass of Ca(OH)2 is 74 g / mol

so 3.47 moles of it will be 256.8 g

So 256.8 grams of calcium hydroxide will be produced.

Now 6 moles of water requires one mole of Ca3P2 so 6.94 moles will require , ( 6.94 / 6 ) = 1.16 moles of Ca3P2

So the excess = 1.23 moles - 1.16 moles = 0.07 moles

0.07 moles of Ca3P2 = 0.07 x 182.18= 12.75 g

so 12.75 g of of Ca3P2 will remain unreacted when the reaction is complete.

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