Chemistry, 19.04.2021 16:10 vhuyrtyy

36. Write a balanced chemical equation for each chemical reaction. (a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide. (b) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas. (c) Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas. (d) Liquid benzene (C6H6) reacts with gaseous oxygen to form carbon dioxide and liquid water.

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Answer from: handog12

a) 2Cu(s) + S(s) → Cu₂S(s)

b) 2SO₂(g) + O₂(g) → 2SO₃(g)

c) 4HCl(aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

d) C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)

Explanation:

First, we have to write the chemical formulas for reactants and products. Then, to identify which is the state of matter of each compound (liquid: l, solid: s, aqueous: aq, gas: g). Finally, we write the equation and balance the atoms.

a) Solid copper reacts with solid sulfur to form solid copper(I) sulfide.

Reactants: solid copper (Cu(s)) and solid sulfur (S(s)). Product: copper(I) sulfide (Cu₂S) because sulfur anion has two negative charges (S⁻²) and we need two copper ions with one positive charge (Cu⁺) to obtain the neutral compound.

The balanced requires a coefficient of 2 for Cu(s) to balance the 2 Cu atoms on the products side:

2Cu(s) + S(s) → Cu₂S(s)

(b) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.

Reactants: sulfur dioxide (SO₂) and oxygen gas (which is a diatomic molecule: O₂). Products: sulfur trioxide gas (SO₃(s)).

The balanced equation is:

2SO₂(g) + O₂(g) → 2SO₃(g)

We need a coefficient of 2 for SO₂ and SO₃ to balance O and S atoms.

(c) Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas.

Reactants: aqueous hydrochloric acid (HCl(aq)) and solid manganese (IV) oxide (MnO₂(s), because we need two oxygen anions O²⁻ to neutralize 4 positive charges in Mn IV). Products: aqueous manganese(II) chloride (MnCl₂, because we need two chloride ions Cl⁻ to neutralize two positive charges in Mn(II)), liquid water (H₂O), and chlorine gas (Cl₂, because it is a diatomic molecule).

The balanced chemical equation is:

4HCl(aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

(d) Liquid benzene (C₆H₆) reacts with gaseous oxygen to form carbon dioxide and liquid water.

Reactants: benzene (C₆H₆) and gaseous oxygen (O₂). Products: carbon dioxide (CO₂, because C has valence IV) and liquid water (H₂O).

The balanced chemical equation is:

C₆H₆(l) + 15/2 O₂(g) → 6CO₂(g) + 3H₂O(l)

In this case, we need a coefficient of 15/2 for O₂ because on the product side there are 15 atoms of O (12 in CO₂ and 3 in H₂O).

Answer from: Quest

Sorry i think the answer is b

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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