The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Step 1: Track yields through the reaction sequence
Initial moles of X = 16 mol
Step 2: Calculate molar mass of S
S is formed from R (via NaH in ether). Given molar mass of P is 210 and the reaction doesn't drastically change the skeleton, so we consider R \( \approx \) P for molar mass estimate. Thus:
\[ \text{Approx. molar mass of S} = 210 \, \text{g/mol} \]
Step 3: Calculate mass of S
\[ n = 2 \, \text{mol}, \quad M = 210 \, \text{g/mol} \Rightarrow \text{Mass of S} = n \cdot M = 2 \times 42 = \boxed{84 \, \text{g}} \]
Final Answer: \( \boxed{84} \)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is