Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Answer:
Magnesium chloride and water
Explanation:
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
magnesium chloride water
Answer:
What procedure did you use to complete the lab?
→Procedures that needs to be considered to complete the lab are- a thorough knowledge of lab assignments, knowledge about safety equipment, reviewing the MSDS of chemicals for lab experiment etc.
<h3>Explanation:</h3>
To be lab prepared one must follow these procedures-
1. One should have the knowledge of lab assignments to make the lab experiment easier.
2. To be aware about safety equipment and their uses in lab, like- the location of fire extinguisher in lab.
3. To know the steps of experiments to be performed
4. To fill notebook of lab with information regarding the experiment
5. One should review the data sheets of chemicals material safety.
6. To put on all the necessary dressings to perform experiment.
7. To have complete understanding about the experiment disposals.
Answer:
number of elements
Explanation:
its the number of elements because I did an educational guess...still appreciate it
