Question

Consider the following unbalanced equation: $${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2(s)+{\mathrm{H}}_2{\mathrm{SO}}_4(aq)⟶{\mathrm{CaSO}}_4(s)+{\mathrm{H}}_3{\mathrm{PO}}_4(aq)$$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 $\mathrm{kg}$ calcium phosphate with 1.0 $\mathrm{kg}$ concentrated sulfuric acid $(98\mathrm{\%}{\mathrm{H}}_2{\mathrm{SO}}_4\text{ by mass)? }$

Short answer

The mass of calcium sulfate produced is $1315.83g$, and the mass of phosphoric acid produced is $631.54g$ when reacting 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% by mass).

Step-by-step solution

Balance the equation.

First, we need to balance the given chemical equation to maintain the conservation of mass: Ca3(PO4)2(s) + H2SO4(aq) → CaSO4(s) + H3PO4(aq) Balanced equation: 2Ca3(PO4)2(s) + 6H2SO4(aq) → 6CaSO4(s) + 4H3PO4(aq)

Calculate the moles of reactants.

Next, we will determine the amounts of reactants available in moles using their respective molar masses: Molar mass of Ca3(PO4)2 = 310.18 g/mol Molar mass of H2SO4 = 98.08 g/mol Given mass of Ca3(PO4)2 = 1.0 kg = 1000 g Given mass of H2SO4 = 1.0 kg × 98 % = 980 g Moles of Ca3(PO4)2 = mass / molar mass Moles of Ca3(PO4)2 = 1000 g / 310.18 g/mol = 3.22 mol Moles of H2SO4 = mass / molar mass Moles of H2SO4 = 980 g / 98.08 g/mol = 9.99 mol

Identify the limiting reactant.

We can now identify the limiting reactant by comparing the ratios of moles of reactants to the stoichiometry of the balanced equation: Mole ratio of Ca3(PO4)2 to H2SO4 in the balanced equation = 2 : 6 Divide the moles of each reactant by their stoichiometric coefficient: Ca3(PO4)2 = 3.22 mol / 2 = 1.61 H2SO4 = 9.99 mol / 6 = 1.67 Since the value for Ca3(PO4)2 is lower, it is the limiting reactant.

Calculate the mass of products formed.

Now, using the limiting reactant (Ca3(PO4)2), we can calculate the mass of the products formed (CaSO4 and H3PO4). Molar mass of CaSO4 = 136.14 g/mol Molar mass of H3PO4 = 97.99 g/mol From the balanced equation, we see that every 2 moles of Ca3(PO4)2 will produce 6 moles of CaSO4. Moles of CaSO4 produced = (3.22 mol Ca3(PO4)2 / 2) × 6 = 9.66 mol Mass of CaSO4 produced = moles × molar mass Mass of CaSO4 = 9.66 mol × 136.14 g/mol = 1315.83 g Similarly, every 2 moles of Ca3(PO4)2 will produce 4 moles of H3PO4. Moles of H3PO4 produced = (3.22 mol Ca3(PO4)2 / 2) × 4 = 6.44 mol Mass of H3PO4 produced = moles × molar mass Mass of H3PO4 = 6.44 mol × 97.99 g/mol = 631.54 g

Present the final answer.

The mass of calcium sulfate produced is 1315.83 g, and the mass of phosphoric acid produced is 631.54 g when reacting 1.0 kg of calcium phosphate with 1.0 kg of concentrated sulfuric acid (98% by mass).

Key concepts

Mole Calculations

To understand mole calculations, begin by realizing that moles are a concept that allows chemists to count particles. The mole connects the mass of a substance to the number of particles it contains.
Consider calcium phosphate $({\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2)$ and sulfuric acid $({\mathrm{H}}_2{\mathrm{SO}}_4)$ in our exercise.

First, we need the molar mass to convert from grams to moles. The molar mass of calcium phosphate is 310.18 g/mol, and for sulfuric acid, it is 98.08 g/mol.
We begin with 1.0 kg of each reactant, which converts to 1000 grams for each.

• For ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1000 \text{ g}}{310.18 \text{ g/mol}} = 3.22 \text{ mol} \
• For ${\mathrm{H}}_2{\mathrm{SO}}_4$: Apply 98% concentration: \ [\text{Mass} = 1.0 \text{ kg} \times 0.98 = 980 \text{ g}\]
  Then, convert to moles:$$\text{Moles}=\frac{980\text{ g}}{98.08\text{ g/mol}}=9.99\text{ mol}$$

Mole calculations are pivotal for understanding quantities in chemical reactions.

Stoichiometry

Stoichiometry is the study and application of quantitative relationships in chemical reactions. It tells us how much of each reactant is needed and how much of each product will form.
Using the balanced chemical equation:
$$2{\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2(s)+6{\mathrm{H}}_2{\mathrm{SO}}_4(aq)\to 6{\mathrm{CaSO}}_4(s)+4{\mathrm{H}}_3{\mathrm{PO}}_4(aq)$$
We see that the stoichiometric ratio of ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$ to ${\mathrm{H}}_2{\mathrm{SO}}_4$ is 2 to 6.
This ratio governs how reactants combine.By dividing the moles of each reactant by its stoichiometric coefficient:

• ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$: $$\frac{3.22}{2}=1.61$$
• ${\mathrm{H}}_2{\mathrm{SO}}_4$: $$\frac{9.99}{6}=1.67$$

From these calculations, we identify ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$ as the limiting reactant because it runs out first.
Therefore, stoichiometry is essential for finding limiting reactants and calculating product masses.

Chemical Equation Balancing

Chemical equation balancing is a crucial step in solving stoichiometry problems.
It ensures the conservation of mass and atoms throughout the chemical process. For our exercise:
$${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2(s)+{\mathrm{H}}_2{\mathrm{SO}}_4(aq)\to {\mathrm{CaSO}}_4(s)+{\mathrm{H}}_3{\mathrm{PO}}_4(aq)$$
is unbalanced.To balance it, adjust coefficients:

• We need 2 moles of ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$.
• React these with 6 moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$.
• This results in 6 moles of ${\mathrm{CaSO}}_4$ and 4 moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$.

Every atom on the left has an equivalent atom on the right.Balancing chemical equations is fundamental for understanding how reactants transform into products and ensuring calculations are correct.

Product Mass Calculation

Product mass calculations allow us to determine the mass of products formed after a chemical reaction.
Start with the information from the limiting reactant.From our balanced equation, each 2 moles of ${\mathrm{Ca}}_3({\mathrm{PO}}_4{)}_2$ produces:

• 6 moles of ${\mathrm{CaSO}}_4$
• 4 moles of ${\mathrm{H}}_3{\mathrm{PO}}_4$

Calculate the mass of ${\mathrm{CaSO}}_4$:

• Moles produced = $$\frac{3.22\text{ mol}}{2}\times 6=9.66\text{ mol}$$
• Mass = $9.66\text{ mol}\times 136.14\text{ g/mol}=1315.83\text{ g}$

Calculate the mass of ${\mathrm{H}}_3{\mathrm{PO}}_4$:

• Moles produced = $$\frac{3.22\text{ mol}}{2}\times 4=6.44\text{ mol}$$
• Mass = $6.44\text{ mol}\times 97.99\text{ g/mol}=631.54\text{ g}$

Product mass calculation leverages mole ratios and molar masses to give detailed insight into the potential outcomes of reactions.