Kumonekta sa mga eksperto at makakuha ng mga sagot sa IDNStudy.com. Ang aming platform ay nagbibigay ng mga maaasahang sagot upang matulungan kang gumawa ng matalinong desisyon nang mabilis at madali.
Sagot :
SOLUTION:
Step 1: Write the balanced chemical equation.
2Al + 6HBr → 3H₂ + 2AlBr₃
Step 2: Calculate the number of moles of H₂ formed by each reactant.
• Using Al
Based on the balanced chemical equation, 2 moles of Al is stoichiometrically equivalent to 3 mole of H₂.
[tex]\begin{aligned} \text{moles of} \: \text{H}_2 & = \text{3.22 mol Al} \times \frac{\text{3 mol} \: \text{H}_2}{\text{2 mol Al}} \\ & = \text{4.83 mol} \end{aligned}[/tex]
• Using HBr
Based on the balanced chemical equation, 6 moles of Hbr is stoichiometrically equivalent to 3 mole of H₂.
[tex]\begin{aligned} \text{moles of} \: \text{H}_2 & = \text{4.96 mol HBr} \times \frac{\text{3 mol} \: \text{H}_2}{\text{6 mol HBr}} \\ & = \text{2.48 mol} \end{aligned}[/tex]
Step 3: Determine the limiting reagent.
Since HBr produced less amount of H₂ than Al, HBr is the limiting reagent.
Step 4: Determine the mass of H₂ formed.
Note that the (maximum) mass of a product is dictated by the limiting reagent. In this case, we will start at the number of moles of H₂ formed from the limiting reagent (HBr) which is equal to 2.48 mol.
The molar mass of H₂ is 2.016 g.
[tex]\begin{aligned} \text{mass of} \: \text{H}_2 & = \text{2.48 mol} \: \text{H}_2 \times \frac{\text{2.016 g} \: \text{H}_2}{\text{1 mol} \: \text{H}_2} \\ & = \boxed{\text{5.00 g}} \end{aligned}[/tex]
Hence, 5.00 g of H₂ are formed.
[tex]\\[/tex]
#CarryOnLearning
Ang iyong presensya ay mahalaga sa amin. Magpatuloy sa pagtatanong at pagbabahagi ng iyong nalalaman. Ang iyong ambag ay napakahalaga sa aming komunidad. Bawat tanong ay may sagot sa IDNStudy.com. Salamat sa pagpili sa amin at sa muling pagkikita.
