This problem is providing the mass of both magnesium metal and oxygen gas and involved in a chemical reaction and asks for the limiting reactant. At the end, it turns out to be identified as magnesium.
<h3>Stoichiometry</h3>
In chemistry, stoichiometry is a widely-used tool we use in order to relate the mass and moles of different chemical substances involved in a chemical reaction. Thus, we consider the following chemical equation between magnesium and oxygen to produce magnesium oxide.
However, when the mass of the both of the reactants is given, one must identify the limiting reactant as the one producing the least of the moles of the product, which means we can use the given grams of the both of the reactants, their molar masses and mole ratios with the product to obtain the aforementioned:
Thus, we can evidence how 24 g of magnesium produce the least of the moles of magnesium oxide, fact validating the magnesium as the limiting reactant and the oxygen as the excess one.
Learn more about stoichiometry: brainly.com/question/9743981
Answer:
- second choice:<em><u> 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene.</u></em>
Explanation:
ppm is a unit of concentration that means parts per million. In grams that is grams of solute per one million (10⁶) grams of solution.
Then, <em>15 ppm of benzene</em> means that there are 15 grams of benzen in 1,000,000 grams of solution.
That leads to:
- 1,000,000 g solution / 15 g benzene
Multiplying numerator and denominator by 10⁻⁶ you find:
- 1,000,000 × 10⁻⁶ g solution / (15 × 10⁻⁶ g benzene)
Simplifying:
- 1.0 g solution / (15 × 10⁻⁶ g benzene)
Which is read as 1.0 g of the solution contains 15 × 10⁻⁶ g of benzene, i.e. the second answer choice.
Answer:
ΔG° = 41.248 KJ/mol (298 K); the correct answer is a) 41 KJ
Explanation:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
⇒ Kf = 1.7 E7; T =298K
⇒ ΔG° = - RT Ln Kf.....for aqueous solutions
∴ R = 8.314 J/mol.K
⇒ ΔG° = - ( 8.314 J/mol.K ) * ( 278 K ) ln ( 1.7 E7 )
⇒ ΔG° = 41248.41 J/mol * ( KJ / 1000J )
⇒ ΔG° = 41.248 KJ/mol
Balanced chemical equation:
* moles of oxygen
4 Al + 3 O2 = 2 Al2O3
4 moles Al -------------- 3 moles O2
9.30 moles Al ---------- moles O2
moles O2 = 9.30 * 3 / 4
moles O2 = 27.9 / 4 => 6.975 moles of O2
Therefore:
Molar mass O2 = 31.9988 g/mol
n = m / mm
6.975 = m / 31.9988
m = 6.975 * 31.9988
m = 223.19 of O2
