Question

How many grams of calcium phosphate can be produced when 89.3 grams of calcium chloride reacts with excess sodium phosphate?

Answer 1

CaCl2+Na3PO4-->Ca3(PO4)2+NaCl
Write down the number of atoms that you have on each side of the equation.
Reactant Side: 1 Calcium, 2 Chlorine, 3 Sodium, 1 Phosphorus, 4 oxygen
Product Side: 3 Calcium, 1 Chlorine, 1 Sodium, 2 Phosphorus, 8 Oxygen
Add a coefficient of 3 to CaCl2
3CaCl2+Na3PO4-->Ca3(PO4)2+NaCl
Reactant Side: 3 Calcium, 6 Chlorine, 3 Sodium, 1 Phosphorus, 4 oxygen
Product Side: 3 Calcium, 1 Chlorine, 1 Sodium, 2 Phosphorus, 8 Oxygen
Add a coefficient of 6 to NaCl and 2 to Na3PO4
3 CaCl2(aq) + 2 Na3PO4(aq) --> Ca3(PO4)2(s) + 6NaCl
Reactant Side: 3 Calcium, 6 Chlorine, 6 Sodium, 2 Phosphorus, 8 oxygen
Product Side: 3 Calcium, 6 Chlorine, 6 Sodium, 2 Phosphorus, 8 Oxygen
Now the chemical equation is balanced.
Double Displacement Reaction Definition: A double displacement reaction is a type of reaction where part of of one reactant is replaced by part of another reactant.
Use Number of moles=Mass(g)/Molar Mass(g/mol)

Answer 2

So, we know that calcium chloride (Chemical Formula: CaCl2) reacts with sodium phosphate (Na3PO4) to produce calcium phosphate (Ca3(PO4)2). So, the calcium ions (Ca2+) from the calcium chloride will come together with the phosphate ions ((PO4)3-) from the sodium phosphate to form the calcium phosphate. Alongside this, the remaining chlorine ion (Cl+) will also come together with the sodium ion (Na+) to form sodium chloride (NaCl). Hence, the chemical equation for the reaction looks like:

\[CaCl _{2} + Na _{3}PO _{4} \rightarrow Ca _{3}(PO _{4})_{2} + NaCl\]

However, to solve this problem we need to look at the BALANCED chemical equation for the reaction. Balancing the above, to make sure there is the same amount of each type of atom on each side of the equation, we get:

\[3CaCl _{2} + 2Na _{3}PO _{4} \rightarrow Ca _{3}(PO _{4})_{2} + 6NaCl\]

We know that the sodium phosphate is in excess, meaning that the calcium chloride is what we refer to as the LIMITING REAGENT. Essentially, the balanced equation above is telling us that for every three moles of CaCl2 we react with every two moles of Na2PO4, we will get 1 mole of Ca3(PO4)2 and 6 moles of NaCl. But we have more of the sodium phosphate than is required to react with the available amount (89.3 grams) of calcium chloride, and so it is the latter which will limit the formation of the products and how much will actually form.

-We need to find out the number of moles of CaCl2 we have. With help form the peridoc table, the relative molecular mass of CaCl2 is:
40 g/mole + 2(35.5 g/mole) = 111 g/mole

We have 89.3 g of CaCl2, so to convert to moles we work out:
\[\frac{ 89.3g }{ 111g/mole } = 0.8045 moles\]

As the CaCl2 is what will determine the amount of calcium phosphate formed, we can use this figure now. The balanced chemical equation tells us that for every 3 moles of CaCl2 we have, if we react them fully (which we do here) we will form 1 mole of Ca3(PO4)2. So, if we have 0.80 moles of the reactant, we will theoretically form 0.80/3 moles of the product, or 0.2681 moles.

-So we now know the number of moles of calcium phosphate we will form. We now need to know the number of grams of the substance this converts. As before, we work out the relative molecular mass of Ca3(PO4)2 to be:
3(40 g/mole) + 2(30 g/mole + 4(16 m/mole)) = 308 g/mole.
To convert we then work out:
\[0.2681moles \times 308 g/mole = ?\] which, if you work out, should give you you're answer. Hope that helps! :)

Answer 3

89.3/110.1=0.811 moles of CaCl2
Look at the mole ratio from the balanced chemical equation. 3 moles of CaCl2 react to produce 1 mole of Ca3(PO4)2. what if 0.811 moles of CaCl2 are used.
3:1
0.811:x mole of Ca3(PO4)2
x=0.811*1/3=0.270 moles
Rearrange the equation Number of moles=Mass(g)/Molar Mass(g/mol)
Look at your periodic table for the elements atomic weight to calculate the molar mass
Mass(g)=Molar Mass(g/mol)*Number of moles
Mass(g)=310.1*0.270=83.8

Answer 4

good job @Ciarán95 :)

Answer 5

Thanks @Abmon98 ! :)