Question

If 16.00 g of O₂ reacts with 80.00 g NO, how many grams of the excess reactant is used in the chemical reaction? (enter only the value, round to whole number)

Answer 1

Answer:

50.00 g of NO

Explanation:

Remember that the balanced chemical reaction equation is indispensable in solving any question that has to do with stoichiometry. Hence the first step in solving the problem is noting down the balanced chemical reaction equation.

2NO(g) + O2 (g)→ 2NO2(g)

Now we try to find out the reactant in excess. The reactant in excess gives the greater mass of product.

For O2;

From the balanced reaction equation;

32 g of O2 yields 92g of NO2

16.00g of O2 will yield 16.00×92/32 = 46g of NO2

For NO;

30g of NO yields 92g of NO2

80.00 g of NO yields 80.00 × 92/30 = 245.33 g of NO2

Hence NO is the reactant in excess.

If 1 mole of O2 reacts with 2 moles of NO2 according to the balanced reaction equation

Then 32 g of O2 reacts with 60g of NO according to the balanced reaction equation

16.00 g of O2 reacts with 16.00 × 60 /32 = 30 g of NO

Hence mass of excess reactant used in the reaction = total mass of NO- mass of NO reacted= 80.00g -30.00g = 50.00 g of NO

Hence the mass of excess reactant used in the reaction is 50.00 g of NO

Answer 2

Answer:

The amount in grams of the excess reactant (NO) used in the chemical reaction is 30 grams

Explanation:

The balanced chemical equation for the reaction is presented as follows;

O₂ (g) + 2NO (g) → 2NO₂ (g)

Therefore, one mole of O₂ combines with two moles of NO to produce two moles NO₂

Molar mass of oxygen gas, O₂ = 31.999 g/mol

Mass of oxygen gas, O₂, present = 16.00 g

Molar mass of nitric oxide, NO = 30.01 g/mol

Mass of nitric oxide, NO, present = 80.008

Number of moles, n, of a sample of a substance is given by the following relation;

$n = \frac{Mass \ m, \, of \, sample}{Molar \, mass \, of \, sample}$

Number of moles of O₂ present is presented as follows;

$n_{O_2}=\frac{16 \, g}{31.999 \, g} = 0.5 \, moles$

Number of moles of NO present is presented as follows;

$n_{NO}=\frac{80 \, g}{30.01 \, g} = 2.67 \, moles$

Hence, since one mole of O₂ combines with two moles of NO to produce two moles NO₂, 0.5 moles of O₂ will combine with one mole of NO to produce one mole NO₂. Therefore, the oxygen is used up in the reaction and the nitric oxide is the excess reactant.

Number of moles of nitric oxide, NO, consumed in the reaction = 1 mole

Mass, m of nitric oxide used = number of moles of nitric oxide × molar mass of nitric oxide

∴ Mass of nitric oxide used = 1 mole × 30.01 g/mol = 30.01 grams

Rounding to the nearest whole number, we have;

Mass of nitric oxide used = 30 grams.

The amount in grams of the excess reactant (NO) used in the chemical reaction = 30 grams.