Question

A solid sample of Rb2SO3 weighing 6.24 g reacts with 1.38 L gaseous HBr, measured at 75°C and 0.953 atm pressure. The solid RbBr, extracted from the reaction mixture and purified, has a mass of 7.32 g.
(a) What is the limiting reactant?
(b) What is the theoretical yield of RbBr, assuming com- plete reaction?
(c) What is the actual percentage yield of product?

Answer 1

Answer:

(a) HBr;

(b) 7.61 g;

(c) 96.2 %

Explanation:

Firstly, write the balanced chemical equation:

$Rb_2SO_3 (aq) + 2 HBr (aq)\rightarrow 2 RbBr (aq) + SO_2 (g) + H_2O (l)$

(a) Find moles of each reactant dividing the mass by the molar mass of rubidium sulfite, then applying the ideal gas law for HBr:

$n_{Rb_2SO_3}=\frac{6.24 g}{251.00 g/mol} = 0.02486 mol$

$pV_{HBr}=n_{HBr}RT$

$\therefore n_{HBr} = \frac{pV_{HBr}}{RT} = \frac{0.953 atm\cdot 1.38 L}{0.08206 \frac{L atm}{mol K}\cdot 348.15 K} = 0.04603 mol$

Find the limiting reactant by dividing each moles by the stoichiometric coefficients and comparing the two numbers:

$eq._{Rb_2SO_3} = \frac{0.02486 mol}{1} = 0.02486 mol$

$eq._{HBr} = \frac{0.04603 mol}{2} = 0.02302 mol$

That said, the equivalent of HBr is lower, so it's the limiting reactant.

(b) According to the balanced equation, the moles of HBr are equal to the moles of RbBr, so moles of RbBr theoretically are equal to:

$n_{RbBr} = 0.04603 mol$

Using the molar mass of RbBr, convert this into mass:

$m_{RbBr} = 0.04603 mol\cdot 165.372 g/mol = 7.61 g$

(c) To find the percent yield, divide the actual mass produced by the theoretical mass calculated in (b) and multiply by 100 %:

$\%_{yield} =\frac{7.32 g}{7.61 g}\cdot 100\% = 96.2 \%$