Answer to Question #137645 in General Chemistry for lyle lasdoce

Question

what mass of nitric oxide canbe made from the reaction of 8.00g NH3 with 17.0g O2

Answer 1

STEP I: BALANCED EQUATION

4NH_3(g) + 5O_2(g) $\to$ 4NO_(g) + 6H2O_(l)

STEP II: MOLES OF EACH REAGENT PRESENT

Moles of ammonia

$Moles =\dfrac{mass}{MM}$

Molar mass of NH₃ is ( N=14, H=1)

MM = 14 +3 = 17g/mol

$Moles =\dfrac{8g}{17g} = 0.471 mol$

Moles of oxygen (MM of O2 = 2 x 16 = 32g/mol)

$Moles =\dfrac{17g}{32g/mol} = 0.25mol$

STEP III: DETERMINE THE LIMITING REAGENT

Moles of oxygen required to react with 0.472mol of ammonia

Mole ratios from the equation (NH3: O2) = 4:5

Thus moles of O2 = $\dfrac{5}{4}x0.471mol$ = 0.589mol.

But there are only 0.25 mol.

Therefore oxygen is the limiting reagent

The amount of product depends on amount of oxygen present.

STEP IV: DETERMINE AMOUNT OF PRODUCT

Moles of NO produced

Mole ratios of O2:NO = 5:4

5 mole of O2 produces 4 moles of NO

0.25 moles of O2 would produce?

Moles of NO $=\dfrac{4}{5}x 0.25mol = 0.2 mol$

Mass of NO = Moles x MM (MM = 14 +16)

Mass of NO = 0.2 mol x 30g/mol

= 6g on NO