What is the maximum amount of Ca3(PO4)2 that can be prepared from 9.80 g of Ca(OH)2 and 9.80 g of H3PO4
Ca(OH)2 (s) + H3PO4 (aq)
Ca3(PO4)2 (aq) + H2O (1)
balance the equation 1st.
O 6.80 g
O 15.5 g
O 8.60 g
o 13.7 g
O 10.3 g

Taking into account the reaction stoichiometry and limiting reagent, 13.68 grams of Ca₃(PO₄)₂ are formed when 9.80 g of Ca(OH)₂ and 9.80 g of  H₃PO₄ react.

Reaction stoichiometry

In first place, the balanced reaction is:

3 Ca(OH)₂  + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Ca(OH)₂: 3 moles  H₃PO₄: 2 moles Ca₃(PO₄)₂: 1 mole H₂O: 6 moles

The molar mass of the compounds is:

Ca(OH)₂: 74 g/mole H₃PO₄: 98 g/moleCa₃(PO₄)₂: 310 g/moleH₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Ca(OH)₂: 3 moles× 74 g/mole= 222 grams H₃PO₄: 2 moles× 98 g/mole= 196 gramsCa₃(PO₄)₂: 1 mole× 310 g/mole= 310 gramsH₂O: 6 moles× 18 g/mole= 108 gramsLimiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction a simple rule of three as follows: if by stoichiometry 196 grams of H₃PO₄ reacts with 222 grams of Ca(OH)₂, if 9.80 grams of H₃PO₄ react with how much mass of Ca(OH)₂ will be needed?

mass of Ca(OH)₂= 11.1 grams

But 11.1 grams of Ca(OH)₂ are not available, 9.80 moles are available. Since you have less moles than you need to react with 9.80 grams of H₃PO₄, Ca(OH)₂ will be the limiting reagent.

Maximum amount of Ca₃(PO₄)₂

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 222 grams of Ca(OH)₂ form 310 grams of Ca₃(PO₄)₂, 9.80 grams of Ca(OH)₂ form how much mass of Ca₃(PO₄)₂?

mass of Ca₃(PO₄)₂= 13.68 grams

Then, 13.68 grams of Ca₃(PO₄)₂ are formed when 9.80 g of Ca(OH)₂ and 9.80 g of  H₃PO₄ react.

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13.7g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

3Ca(OH)2(s) + 2H3PO4(aq) —> Ca3(PO4)2(aq) + 12H2O(l)

Step 2:

Determination of the masses of Ca(OH)2 and H3PO4 that reacted and the mass of Ca3(PO4)2 produced from the balanced equation.

Molar mass of Ca(OH)2 = 40 + 2(16 + 1) = 74g/mol

Mass of Ca(OH)2 from the balanced equation = 3 x 74 = 222g

Molar mass of H3PO4 = (3x1) + 31 + (16x4) = 98g/mol

Mass of H3PO4 from the balanced equation = 2 x 98 = 196g

Molar mass of Ca3(PO4)2 = (40x3) + 2[31 + (16x4)]

= 120 + 2[95] = 310g/mol

Mass of Ca3(PO4)2 from the balanced equation = 1 x 310 = 310g.

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4 to produce 310g of Ca3(PO4)2.

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4.

Therefore, 9.8g of Ca(OH)2 will react with = (9.8 x 196)/222 = 8.65g of H3PO4

From the above illustration, we can see that only 8.65g of H3PO4 out 9.8g given reacted completely with 9.8g of Ca(OH)2. Therefore, Ca(OH)2 is the limiting reactant and H3PO4 is the excess reactant.

Step 4:

Determination of the maximum mass of Ca3(PO4)2 produced from reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction.

The limiting reactant is Ca(OH)2 and maximum mass of Ca3(PO4)2 produced can be obtained as follow:

From the balanced equation above,

222g of Ca(OH)2 reacted to produce 310g of Ca3(PO4)2.

Therefore, 9.8g of Ca(OH)2 will react to produce = (9.8 x 310)/222 = 13.7g of Ca3(PO4)2.

Therefore, the maximum mass of Ca3(PO4)2 produced is 13.7g

Measure the mass of the platinum before and after the reaction 
as if it is acting as a catalyst the mass of platinum will remain unchanged as catalyst do not consume themselves in reaction