Some rocket engines use a mixture of hydrazine, `N_(2)H_(4)` and hydrogen peroxide, `H_2O_2` as the propellant. The reaction is given by the following equation :
`N_(2)H_(4)(l) + 2H_(2)O_(2)(l) to N_(2)(g) + 4H_(2)O(g)`
How much of the excess reactant, remains unchanged when 0.750 mol of `N_2H_4` is mixed with 17 g of `H_(2)O_(2)` ?

Options

A. 16 g `N_(2)H_(4)`
B. 0.25 mol `H_(2)O_(2)`
C. 0.25 mol `N_(2)H_(4)`
D. 8.5 g `H_(2)O_(2)`

Answer

A

Explanation

`N_(2)H_(4)(l) + 2H_(2)O_(2)(l) to N_(2)(g) + 4H_(2)O (g)`
2 moles of `H_2O_2` require 1 mole of `N_2H_4`.
`therefore` 0.5 mole of `H_(2)O_(2)` will require `=1/2 xx 0.5` mole of `N_2H_4`
`therefore` Moles of `N_2H_4` unreacted = 0.75 - 0.25 = 0.5 mole
Mass of `N_2H_4` unreacted = `0.5 xx 32 = 16 g`