Question

1. if 98.0 g b2o3 (69.62 g/mol) and 125 g hf (20.01 g/mol) are combined and allowed to react according to the given equation, what mass of the excess reactant will be left over?

b₂o₃(s) + 6 hf(aq) → 2 bf₃(g) + 3 h₂o(ℓ)

a. 72.5 g b2o3 will be left over.
b. 25.5 g b2o3 will be left over.
c. 28.2 g hf will be left over.
d. 7.34 g hf will be left over.
e. both are completely consumed.

1. metallic chromium can be obtained from the mineral chromite (fecr2o4). what is the mass percent of iron in chromite?

fecr₂o₄ =
a. 46.46%
b. 61.90%
c. 24.97%
d. 30.26%
e. 41.99%

1. the dissociation of weak electrolytes is different than the dissociation of strong electrolytes because weak electrolytes __________ and the reaction is _________.

i. dissociate completely
ii. dissociate partially
iii. reversible
iv. irreversible

a) i and iii
b) ii and iii
c) i and iv
d) ii and iv

Explanation:

Step1: Calculate the moles of each reactant

• Moles of \(B_{2}O_{3}\): \(n_{B_{2}O_{3}}=\frac{m}{M}=\frac{98.0\ g}{69.62\ g/mol}\approx1.41\ mol\)
• Moles of \(HF\): \(n_{HF}=\frac{m}{M}=\frac{125\ g}{20.01\ g/mol}\approx6.25\ mol\)

Step2: Determine the limiting reactant

From the balanced equation \(B_{2}O_{3}(s)+6HF(aq)\to2BF_{3}(g)+3H_{2}O(l)\), the mole ratio of \(B_{2}O_{3}\) to \(HF\) is \(1:6\).
If all \(B_{2}O_{3}\) reacts, it needs \(n_{HF\ required}=6\times1.41\ mol = 8.46\ mol\). But we have only \(6.25\ mol\) of \(HF\).
If all \(HF\) reacts, it needs \(n_{B_{2}O_{3}\ required}=\frac{6.25\ mol}{6}\approx1.04\ mol\)

So \(HF\) is the limiting reactant.

Step3: Calculate the moles of \(B_{2}O_{3}\) reacted

Since \(HF\) is limiting, moles of \(B_{2}O_{3}\) reacted \(n_{B_{2}O_{3}\ reacted}=\frac{6.25\ mol}{6}\approx1.04\ mol\)

Step4: Calculate the moles of \(B_{2}O_{3}\) left

Moles of \(B_{2}O_{3}\) left \(n_{B_{2}O_{3}\ left}=1.41\ mol - 1.04\ mol=0.37\ mol\)

Step5: Calculate the mass of \(B_{2}O_{3}\) left

Mass of \(B_{2}O_{3}\) left \(m = n\times M=0.37\ mol\times69.62\ g/mol\approx25.7\ g\approx25.5\ g\)

Step1: Calculate the molar mass of \(FeCr_{2}O_{4}\)

• Molar mass of \(Fe = 55.85\ g/mol\)
• Molar mass of \(Cr = 52.00\ g/mol\)
• Molar mass of \(O = 16.00\ g/mol\)

\(M_{FeCr_{2}O_{4}}=55.85+(2\times52.00)+(4\times16.00)=55.85 + 104+64=223.85\ g/mol\)

Step2: Calculate the mass percent of \(Fe\)

Mass percent of \(Fe=\frac{55.85\ g/mol}{223.85\ g/mol}\times100\%\approx24.97\%\)

Weak electrolytes dissociate partially (\(II\)) and their dissociation reaction is reversible (\(III\)) because there is an equilibrium between the undissociated and dissociated forms. Strong electrolytes dissociate completely (\(I\)) and the reaction is irreversible (\(IV\)) as they fully break into ions.

Answer:

b. \(25.5\ g\ B2O3\) will be left over.