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:

Explanation:
First, we need to find the molecular mass of water (H₂O).
H₂O has:
- 2 Hydrogen atoms (subscript of 2)
- 1 Oxygen atom (implied subscript of 1)
Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.
- Hydrogen: 1.0079 g/mol
- Oxygen: 15.9994 g/mol
There are 2 hydrogen atoms, so multiply the mass by 2.
- 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol
Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.
- 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol
Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

Multiply. The grams of H₂O will cancel out.



The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.
The 7 in the hundred thousandth place tells us to round up.

There are about <u>0.0018 moles in 0.032 grams.</u>
Explanation:
B. HCIO these is the answer
Answer:
sodium atoms are much larger than chlorine atoms but in case of sodium ions they are much smaller thqn chlorine ions