Tembaga (II) sulfida digunakan dalam industri sebagai bahan dalam semikonduktor, katalis, serta pigmen untuk pewarnaan kaca dan keramik. Bu Ani ingin membuat tembaga (II) sulfida dengan mereaksikan 10 gram tembaga dan 20 gram belerang dengan reaksi, \( \mathrm{Cu}_{(\mathrm{s})}+\mathrm{S}_{(\mathrm{s})} \rightarrow \mathrm{CuS}_{(\mathrm{s})} \) Jika \( \mathrm{Ar} \mathrm{Cu}=63,5 \) dan \( \mathrm{S}=32 \), massa CuS yang terbentuk adalah \( \ldots \). A. 14,99 gram B. 15,99 gram C. 16,99 gram D. 17,99 gram E. 18,99 gram

1. The reaction is given by \[ \mathrm{Cu}_{(\mathrm{s})} + \mathrm{S}_{(\mathrm{s})} \rightarrow \mathrm{CuS}_{(\mathrm{s})} \] 2. The atomic masses are: \[ \mathrm{Ar}_{\mathrm{Cu}} = 63.5, \quad \mathrm{Ar}_{\mathrm{S}} = 32 \] Thus, the molar mass of \(\mathrm{CuS}\) is \[ M_{\mathrm{CuS}} = 63.5 + 32 = 95.5 \text{ g/mol} \] 3. Calculate the number of moles of each reactant: - For \(\mathrm{Cu}\): \[ n_{\mathrm{Cu}} = \frac{10 \text{ g}}{63.5 \text{ g/mol}} \approx 0.1575 \text{ mol} \] - For \(\mathrm{S}\): \[ n_{\mathrm{S}} = \frac{20 \text{ g}}{32 \text{ g/mol}} = 0.625 \text{ mol} \] 4. Compare moles based on the 1:1 stoichiometry in the reaction. Since \[ n_{\mathrm{Cu}} (0.1575 \text{ mol}) < n_{\mathrm{S}} (0.625 \text{ mol}), \] copper is the limiting reagent. 5. The moles of \(\mathrm{CuS}\) formed are equal to the moles of \(\mathrm{Cu}\) (the limiting reagent): \[ n_{\mathrm{CuS}} = 0.1575 \text{ mol} \] 6. Calculate the mass of \(\mathrm{CuS}\) produced: \[ m_{\mathrm{CuS}} = n_{\mathrm{CuS}} \times M_{\mathrm{CuS}} = 0.1575 \text{ mol} \times 95.5 \text{ g/mol} \approx 15.04 \text{ g} \] Rounding to a value close to the provided option, the mass of \(\mathrm{CuS}\) is approximately \(14.99\) grams. The correct answer is A. 14,99 gram.