Question

Срочно! When 5.4 g of aluminum Al reacts with oxygen 10.2 g of aluminum oxide Al2O3 is formed. What is the mass of oxygen reacted. a) write a chemical reaction here: b) find n(Al) = mol c) find n(Al2O3) = mol d) mass of oxygen reacted = g

Answer 1

Ответ:

Answer 4.8 g

а) 2*26.98 = 53.96

b) 3*16 = 48

48 + 53.96 = 101.96

c) m O2 = 10.2 * 48 / 101.96 = 4.8 g

Answer 2

a) The chemical reaction can be represented as follows:

4 Al + 3 O2 → 2 Al2O3

b) To find the number of moles of aluminum (n(Al)), we use the molar mass of aluminum (Al), which is approximately 26.98 g/mol.

Given mass of aluminum (m(Al)) = 5.4 g

n(Al) = m(Al) / M(Al) n(Al) = 5.4 g / 26.98 g/mol ≈ 0.2001 mol (rounded to four decimal places)

c) According to the balanced chemical equation, the stoichiometric ratio between aluminum (Al) and aluminum oxide (Al2O3) is 4:2, which means for every 4 moles of aluminum, 2 moles of aluminum oxide are formed.

n(Al2O3) = (n(Al) / 4) * 2 n(Al2O3) = (0.2001 mol / 4) * 2 ≈ 0.1000 mol (rounded to four decimal places)

d) To find the mass of oxygen reacted, we can subtract the mass of aluminum oxide (Al2O3) formed from the total mass of aluminum and oxygen reactants.

Mass of oxygen reacted = Total mass of reactants - Mass of aluminum oxide Mass of oxygen reacted = (5.4 g + x g) - 10.2 g (where x represents the mass of oxygen)

Simplifying the equation, we get:

x = Mass of oxygen reacted x = (5.4 g + x g) - 10.2 g x - x = 5.4 g - 10.2 g 0 = -4.8 g

Since we obtained a negative value, it suggests that there might be an error in the given data or calculations. Please double-check the values provided.