The balanced combustion reaction of glucose is:
\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}.\]
From the equation:
1 mol of glucose requires 6 mol of $\text{O}_2$.
Molar mass of glucose = $180 \, \text{g/mol}$.
Molar mass of $\text{O}_2 = 32 \, \text{g/mol}$.
Number of moles of glucose in $900 \, \text{g}$:
\[n = \frac{900}{180} = 5 \, \text{mol}.\]
Oxygen required:
\[\text{Mass of } \text{O}_2 = 5 \cdot 6 \cdot 32 = 960 \, \text{g}.\]
Final Answer:
$960 \, \text{g}$.
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32