Answer:
1.04 moles of AlCl₃
Explanation:
Balanced Equation: 2Al°(s) + 3Cl₂(g) => 2AlCl₃(s)
Approach => convert given data (28 g Al°) to moles. The number of moles AlCl₃ would equal the number of moles aluminum because coefficients for Aluminum and Aluminum Chloride are equal.
moles Al° = (28g)/(27g/mole ) = 1.04 mole Al°
From balanced equation moles AlCl₃ produced equals moles of Al° produced = 1.04 mole AlCl₃ because coefficients of Al° & AlCl₃.
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Answer:
<h2>6.75 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 2.7 × 2.5
We have the final answer as
<h3>6.75 g</h3>
Hope this helps you
<h3><u>Answer;</u></h3>
<u>= 128.772 g of water</u>
2C6H14 + 13 O2 → 6CO2 + 14 H2O
<h3><u>Explanation</u>;</h3>
1 mole of Hexane contains 86 g
Therefore;
87.91 grams of hexane will contain
= 87.91 g/86 g
= 1.022 moles
The balanced reaction for the combustion of Hexane is given by
2C6H14 + 13 O2 → 6CO2 + 14 H2O
Therefore; the mole ratio of C6H14 :H2O is
= 2 : 14
= 1: 7
therefore moles of water from 1.022 moles of Hexane will be;
= 1.022 ×7
= 7.154 moles
Mass of water will be;
= 7.154 moles× 18
<u>= 128.772 g of water</u>
The answer is B) dry and irregular
