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klemol [59]
3 years ago
12

3. Complete the following table.

Chemistry
1 answer:
Vinil7 [7]3 years ago
5 0
Li and CI or C and H
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Which chemical equation demonstrates the law of conservation of mass!
zlopas [31]

Answer:

F. 2NO + 02 —> 2NO

H. 4NH3 + 502 —> 4NO + 6H20

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.

2NO + 02 —> 2NO

From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.

4NH3 + 502 —> 4NO + 6H20

From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.

The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.

This is certified by the law of conservation of mass.

The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side

5 0
3 years ago
1.) Rust forms when Fe, O₂, and H₂O react in the balanced equation below;
valentinak56 [21]

Answer:

1.) A.) The limiting reactant is Fe.

B.) 16.17 g.

2.) 84.70 %.

Explanation:

For the balanced equation:

<em>2Fe(s) + O₂(g) + 2H₂O(l) → 2Fe(OH)₂(s).</em>

2.0 moles of Fe reacts with 1.0 mole of oxygen and 2.0 moles of water to produce 2.0 moles of Fe(OH)₂.

<em>A.) Which of these reactants is the limiting reagent? </em>

  • To determine the limiting reactant, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass.
  • Suppose that water is exist in excess.

no. of moles Fe = mass/atomic mass = (10.0 g)/(55.845 g/mol) = 0.179 mol ≅ 0.18 mol.

no. of moles of O₂ = mass/molar mass = (4.0 g)/(32.0 g/mol) = 0.125 mol.

  • Since from the balanced equation; every 2.0 moles of Fe reacts with 1.0 mole of oxygen.

<em>So, 0.18 mol of Fe reacts with 0.09 mol of O₂.</em>

<em>Thus, the limiting reactant is Fe.</em>

<em>The reactant in excess is O₂ (0.125 mol - 0.09 mol = 0.035 mol).</em>

<em>B.) How many grams of Fe(OH)₂ are formed?</em>

<em><u>Using cross multiplication:</u></em>

∵ 2.0 moles of Fe produce → 2.0 moles of Fe(OH)₂.

∴ 0.18 moles of Fe produce → 0.18 moles of Fe(OH)₂.

∴ The mass (no. of grams) of produced 0.18 mol of Fe(OH)₂ = no. of moles x molar mass = (0.18 mol)(89.86 g/mol) = 16.17 g.

<em>2.) (Using the reaction listed in question 1.) If 2.00 g Fe is reacted with an excess of O₂ and H₂0, and a total of 2.74 g of Fe(OH)₂ is actually obtained, what is the % yield?</em>

The % yield = [(actual mass/calculated mass)] x 100.

The actual mass = 2.74 g.

  • We need to calculate the theoretical mass:

Firstly, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass.

no. of moles Fe = mass/atomic mass = (2.0 g)/(55.845 g/mol) = 0.0358 mol ≅ 0.036 mol.

<em><u>Using cross multiplication:</u></em>

∵ 2.0 moles of Fe produce → 2.0 moles of Fe(OH)₂.

∴ 0.036 moles of Fe produce → 0.036 moles of Fe(OH)₂.

<em>∴ The calculated mass (no. of grams) of produced 0.036 mol of Fe(OH)₂ = no. of moles x molar mass</em> = (0.036 mol)(89.86 g/mol) = <em>3.235 g.</em>

<em>∴ The % yield = [(actual mass/calculated mass)] x 100</em> = [(2.74 g/3.235 g)] x 100 = <em>84.70 %.</em>

4 0
3 years ago
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