Stoikiometri Reaksi Kalium dengan Asam Klorida

ChemistryStoichiometry and Ideal Gas LawMedium

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3. SOAL STANDAR SBMPTN. Sebanyak 3,9 gram kalium direaksikan dengan asam klorida berlebih di dalam ruang yang memiliki suhu 27°C dan tekanan 1 atm, maka volume gas $H_2$ yang terbentuk adalah.... (diketahui $R = 0,082$ L atm mol$^{-1}$K$^{-1}$, Ar K = 39) A. 4,92 L B. 1,83 L C. 1,23 L D. 4,92 mL E. 1,23 mL

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1
Step 1

Halo Aisyah, mari kita selesaikan soal stoikiometri gas ini. Pertama, kita catat data yang diketahui dari soal.

Data Diketahui

$$\begin{aligned} m_{K} &= 3,9\text{ gram} \\ T &= 27^{\circ}\text{C} \\ P &= 1\text{ atm} \\ R &= 0,082\text{ L atm/mol K} \\ Ar_{K} &= 39 \end{aligned}$$
2
Step 2

Langkah pertama adalah menuliskan persamaan reaksi yang setara antara logam kalium dengan asam klorida cair.

1. Persamaan Reaksi

$$2K_{(s)} + 2HCl_{(aq)} \rightarrow 2KCl_{(aq)} + H_{2(g)}$$
3
Step 3

Selanjutnya, kita hitung jumlah mol kalium yang bereaksi menggunakan rumus massa dibagi massa atom relatif atau Ar.

2. Menghitung Mol Kalium

$$n_{K} = \frac{m}{Ar} = \frac{3,9}{39} = 0,1\text{ mol}$$
4
Step 4

Berdasarkan koefisien reaksi, kita bisa menentukan mol gas hidrogen. Perbandingan koefisien hidrogen terhadap kalium adalah satu banding dua.

3. Menghitung Mol Gas Hidrogen

$$n_{H_2} = \frac{1}{2} \times n_{K} = \frac{1}{2} \times 0,1 = 0,05\text{ mol}$$
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Step 5

Sekarang, sebelum menghitung volume, kita harus mengubah suhu ke satuan Kelvin dengan menambahkan dua ratus tujuh puluh tiga.

Konversi Suhu dan Gas Ideal

$$T = 27 + 273 = 300\text{ K}$$

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About This Question

Subject
Chemistry
Topic
Stoichiometry and Ideal Gas Law
Difficulty
Medium
Question Type
Multiple Choice

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