Question

1. phosphoric acid can be prepared by reaction of sulfuric acid, h₂so₄, with calcium phosphate, ca₃(po₄)₂, according to the following equation: ca₃(po₄)₂ + 3h₂so₄ → 3caso₄ + 2h₃po₄ which substance is the limiting reactant when 103.0 g of ca₃(po₄)₂ reacts with and 75.0g of h₂so₄? (ca₃(po₄)₂ = 310.2 g/mol; h₂so₄ = 98.06 g/mol) a. ca₃(po₄)₂ b. h₂so₄ c. caso₄ d. h₃po₄ e. none of the above

Explanation:

Step1: Calculate the moles of \(Ca_3(PO_4)_2\)

Using the formula \(n=\frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass.
For \(Ca_3(PO_4)_2\), \(n_{Ca_3(PO_4)_2}=\frac{103.0\ g}{310.2\ g/mol}\approx0.332\ mol\)

Step2: Calculate the moles of \(H_2SO_4\)

For \(H_2SO_4\), \(n_{H_2SO_4}=\frac{75.0\ g}{98.06\ g/mol}\approx0.765\ mol\)

Step3: Determine the mole ratio from the balanced equation

From \(Ca_3(PO_4)_2 + 3H_2SO_4
ightarrow3CaSO_4 + 2H_3PO_4\), the mole ratio of \(Ca_3(PO_4)_2\) to \(H_2SO_4\) is \(1:3\)

Step4: Calculate the moles of \(H_2SO_4\) required if \(Ca_3(PO_4)_2\) is completely consumed

If \(n_{Ca_3(PO_4)_2} = 0.332\ mol\), then the moles of \(H_2SO_4\) required \(n_{H_2SO_4\ (required)}=3\times n_{Ca_3(PO_4)_2}=3\times0.332\ mol = 0.996\ mol\)

Step5: Compare the available moles of \(H_2SO_4\) with the required moles

We have \(n_{H_2SO_4\ (available)} = 0.765\ mol\) and \(n_{H_2SO_4\ (required)}=0.996\ mol\). Since \(0.765\ mol<0.996\ mol\), \(H_2SO_4\) will be completely consumed first.

Answer:

B. \(H_2SO_4\)