Asolid 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?

(a) HBr;

(b) 7.61 g;

(c) 96.2 %

Explanation:

Firstly, write the balanced chemical equation:

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

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

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:

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

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

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol