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:
$960 \, \text{g}$.
X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X =_____ g. (nearest integer) [Given : molar mass (in g mol\(^{-1}\)) C : 12, H : 1, O : 16, N : 14]
Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Match List-I with List-II: List-I