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:
$960 \, \text{g}$.
The values of X, Y, and Z in the following chemical equation are respectively:
\( S_8 + X HNO_3 ({conc.}) \rightarrow Y H_2SO_4 + X NO_2 + Z H_2O \)
LIST I | LIST II |
---|---|
A. 88 g of CO2 | I. 2 mol |
B. 6.023 × 1023 molecules of H2O | II. 1 mol |
C. 96 g of O2 | III. 6.023 × 1023 molecules |
D. 1 mol of any gas | IV. 3 mol |