Question

how many grams of carbon monoxide are needed to react completely with 16 grams of oxygen gas? 2 CO (g) + O2(g) → 2 CO2 (g)

Answer 1

To react completely with 16 grams of oxygen gas, 28 grams of carbon monoxide is needed following the stoichiometry of the given balanced chemical equation 2 CO(g) + O2(g) → 2 CO2(g).

Step-by-step explanation:

To find out how many grams of carbon monoxide are needed to react completely with 16 grams of oxygen gas, we start by looking at the balanced chemical equation 2 CO(g) + O2(g) → 2 CO2(g). According to this equation, 2 moles of carbon monoxide react with 1 mole of oxygen gas to produce 2 moles of carbon dioxide.

First, we need to calculate the number of moles of oxygen in 16 grams. Since the molar mass of O2 is 32 g/mol (16 g/mol × 2), 16 grams of O2 equals 0.5 moles (16 g ÷ 32 g/mol). The stoichiometry of the reaction tells us that 2 moles of CO are needed for every 1 mole of O2, hence we need 1 mole of CO to react with 0.5 moles of O2.

Now, since the molar mass of CO is 28 g/mol (12 g/mol for C + 16 g/mol for O), 1 mole of CO weighs 28 grams. Therefore, 1 mole or 28 grams of CO is required to react completely with 16 grams of oxygen gas.

Answer 2

Molar mass :

O2 = 31.99 g/mol
CO = 28.01 g/mol

2 CO (g) + O2(g) = 2 CO2 (g)

2 x 28.01 g CO -------------------- 31.99 g O2
mass CO --------------------------- 16 g O2

mass CO = 16 x 2 x 28.01 / 31.99

mass CO = 896.32 / 31.99

=> 28.018 g of CO