To determine how many grams of HBr are required to react with 2 g of propyne (C₃H₄), we need to follow a series of steps involving stoichiometry, which is the calculation of reactants and products in chemical reactions. ### Step 1: Write the Balanced Chemical Equation Propyne (C₃H₄) reacts with hydrogen bromide (HBr) in an addition reaction. The balanced chemical equation for the reaction is: \[ \text{C}_3\text{H}_4 + \text{HBr} \rightarrow \text{C}_3\text{H}_5\text{Br} \] This equation shows that one mole of propyne reacts with one mole of HBr. ### Step 2: Calculate the Molar Mass of Propyne To find out how many moles of propyne we have in 2 g, we first need to calculate the molar mass of propyne (C₃H₄): - Carbon (C): 12 g/mol × 3 = 36 g/mol - Hydrogen (H): 1 g/mol × 4 = 4 g/mol Adding these together gives: \[ \text{Molar mass of C}_3\text{H}_4 = 36 \, \text{g/mol} + 4 \, \text{g/mol} = 40 \, \text{g/mol} \] ### Step 3: Calculate the Number of Moles of Propyne Now, we can calculate the number of moles of propyne in 2 g: \[ \text{Number of moles of C}_3\text{H}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{40 \, \text{g/mol}} = 0.05 \, \text{mol} \] ### Step 4: Determine the Moles of HBr Required From the balanced equation, we see that 1 mole of propyne reacts with 1 mole of HBr. Therefore, the number of moles of HBr required is the same as the number of moles of propyne: \[ \text{Moles of HBr required} = 0.05 \, \text{mol} \] ### Step 5: Calculate the Molar Mass of HBr Next, we need to calculate the molar mass of HBr: - Hydrogen (H): 1 g/mol - Bromine (Br): 80 g/mol Adding these together gives: \[ \text{Molar mass of HBr} = 1 \, \text{g/mol} + 80 \, \text{g/mol} = 81 \, \text{g/mol} \] ### Step 6: Calculate the Mass of HBr Required Now we can calculate the mass of HBr required using the number of moles we found: \[ \text{Mass of HBr} = \text{moles} \times \text{molar mass} = 0.05 \, \text{mol} \times 81 \, \text{g/mol} = 4.05 \, \text{g} \] ### Step 7: Rounding to the Nearest Option Since the options provided are in whole grams, we can round 4.05 g to 4.1 g. However, since the closest option is 4.1 g, we can consider it as 4.1 g, which is not listed in the options. The closest option to our calculation is **A. 4.1 g**. ### Explanation of Other Options - **Option B (6.1 g)**: This is too high based on our calculations. It suggests that more moles of HBr are needed than what is actually required. - **Option C (8.1 g)**: This is also too high and does not match our calculated requirement. - **Option D (16.2 g)**: This is significantly higher than what is needed and indicates a misunderstanding of the stoichiometry involved. ### Revision Summary - **Balanced Equation**: C₃H₄ + HBr → C₃H₅Br shows a 1:1 mole ratio. - **Molar Mass Calculation**: Propyne (C₃H₄) = 40 g/mol; HBr = 81 g/mol. - **Moles Calculation**: 2 g of propyne = 0.05 mol. - **Mass of HBr Required**: 0.05 mol × 81 g/mol = 4.05 g, rounded to 4.1 g. In conclusion, the correct answer is **A. 4.1 g**.