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>
The answer should be D all of the above
Answer:
10 L of CO₂.
Explanation:
The balanced equation for the reaction is given below:
2CO + O₂ —> 2CO₂
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Therefore, 10 L of CO will also react to produce 10 L of CO₂.
Thus, 10 L of CO₂ were obtained from the reaction.
Density=mass/ volume so you solve for volume and get 461.96 mL
Answer: 207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu
2.855 + 49.639 + 45.741 + 108.979 = <em><u>207.217amu</u></em>
Explanation:
