Question

How many grams of oxygen are needed to react completely with 200.0 g of ammonia, nh3? 4nh3(g) + 5o2(g) ? 4no(g) + 6h2o(g)?

Answer 1

The balanced chemical reaction is:

4NH3(g) + 5O2(g) = 4NO(g) + 6H2O

We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows:

200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed

Answer 2

Answer : The mass of oxygen needed are 470.592 grams.

Solution : Given,

Mass of $NH_3$ = 200.0 g

Molar mass of $NH_3$ = 17 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $NH_3$.

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{200.0g}{17g/mole}=11.765moles$

Now we have to calculate the moles of $O_2$  

The balanced chemical reaction is,

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

From the reaction, we conclude that

As, 4 mole of $NH_3$ react with 5 mole of $O_2$

So, 11.765 moles of $NH_3$ react with $\frac{11.765}{4}\times 5=14.706$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(14.706moles)\times (32g/mole)=470.592g$

Therefore, the mass of oxygen needed are 470.592 grams.