Question

For each of the following unbalanced reactions, suppose exactly 1.48 g of each reactant is taken. Indicate which reactant is the limiting reagent. Calculate the mass of each product that is expected.

(a) UO₂(s) + HF(aq) → UF₄(aq) + H₂O(l)
limiting reagent

[_]UO₂

[_]HF   

UF4 produced	g
H2O produced	g

(b) NaNO₃(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + HNO₃(aq)
limiting reagent

[_] NaNO₃

[_] H₂SO₄   

Na2SO4 produced	g
HNO3 produced	g

(c) Zn(s) + HCl(aq) → ZnCl₂(aq) + H₂(g)
limiting reagent

[_] Zn

[_] HCl   

ZnCl2 produced	g
H2 produced	g

(d) B(OH)₃(s) + CH₃OH(l) → B(OCH₃)₃(s) + H₂O(l)
limiting reagent

[_] B(OH)₃

[_] CH₃OH   

B(OCH3)3 produced	g
H2O produced	g

Answer 1

(a)

Balanced equation: UO₂(s) + 4 HF(aq) → UF₄(aq) + 2 H₂O(l)

moles UO2 = 1.48 g / 270.02771 g/mol = 0.00548091897 mol

moles HF = 1.48g / 20.006343 g/mol = 0.07397653834 mol

limiting reagent = UO₂

mass UF4 = 0.00548091897 mol x 314.022523 g/mol = 1.72113200332

UF₄ produced = 1.72 g

H₂O produced = 0.197 g

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(b)

Balanced equation: 2 NaNO₃(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 HNO₃(aq)

moles NaNO3 = 1.48 g / 84.9947 g/mol = 0.01741285044 mol

moles H2SO4 = 1.48g / 98.0785 g/mol = 0.01508995345 mol

limiting reagent = NaNO₃

mass Na₂SO₄ = (1/2) x 0.01741285044 mol x 142.0421 g/mol = 1.23667892174 g

Na₂SO₄ produced = 1.24 g

HNO₃ produced = 1.10 g

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(c)

Balanced equation: Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)

moles Zn = 1.48 g / 65.3800 g/mol = 0.02263689201 mol

moles HCl = 1.48g / 36.4609 g/mol = 0.04059142807 mol

limiting reagent = HCl

mass ZnCl₂ = 0.02029571403 mol x 136.2860 g/mol = 2.76602168229 g

ZnCl₂ produced = 2.77 g

H₂ produced = 0.0409 g

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(d)

Balanced equation: B(OH)₃(s) + 3 CH₃OH(l) → B(OCH₃)₃(s) + 3 H₂O(l)

moles B(OH)₃ = 1.48 g / 61.8330 g/mol = 0.023935439 mol

moles CH₃OH = 1.48g / 32.0419 g/mol = 0.04618952059 mol

limiting reagent = CH₃OH

mass B(OCH₃)₃ = 0.01539650686 mol x 103.9128 g/mol = 1.59989413804 g

B(OCH₃)₃ produced = 1.60 g

H₂O produced = 0.832 g

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