Reaction container holds 5.77g of P4 and 5.77g of O2. The following reaction occurs: P4 + O2 → P4O6.

If enough oxygen is available then the P4O6 reacts further:

P4O6 + O2 → P4O10. (P=31, O =16)

a. What is the limiting reagent for the formation of P4O10? (1mk)

b. What mass of P4O10 is produced? (3mks)

c. What mass of excess reactant is left in the reaction container? (1reaction container holds 5.77g of P4 and 5.77g of O2. The following reaction occurs:

P4 + O2 → P4O6.

If enough oxygen is available then the P4O6 reacts further:

P4O6 + O2 → P4O10. (P=31, O =16)

a. What is the limiting reagent for the formation of P4O10? (1mk)

b. What mass of P4O10 is produced? (3mks)

c. What mass of excess reactant is left in the reaction container? (1

a. O₂ is limiting reactant

b. 5.68g P₄O₁₀ are produced

c. 5.83g P₄O₆ are left in the reaction container

Explanation:

Based in the first reaction,

P₄ + 3O₂ → P₄O₆

1 mole of P₄ reacts with 3 moles of oxygen

Initial moles of P₄ and O₂ are:

Moles P₄ (Molar mass: 124g/mol):

5.77g P₄ * (1mol / 124g) = 0.0465 moles P₄

Moles O₂ (32g/mol):

5.77g O₂ * (1mol / 32g) = 0.180 moles O₂

For a complete reaction of 0.0465 moles P₄ are required:

0.0465 moles P₄ * (3 moles O₂ / 1 mol P₄) = 0.140 moles of O₂

That means will remain 0.040 moles of O₂ and there are produced 0.0465 moles of P₄O₆

For the second reaction:

P₄O₆ + 2O₂ → P₄O₁₀

2 moles of oxygen reacts per mole of P₄O₆

For a complete reaction of P₄O₆ are required:

0.0465 moles P₄O₆ * (2 moles O₂ / 1mol P₄O₆) = 0.093 moles of O₂. As there are just 0.040 moles of O₂,

b. There are produced:

0.040 moles O₂ * (1 mole P₄O₁₀ / 2 moles O₂) = 0.020 moles P₄O₁₀

In mass (Molar mass: 284g/mol):

0.020 moles * (284g / mol) =

c. Also, 0.020 moles P₄O₆ are reacting and will remain:

0.0465 mol - 0.020 mol = 0.0265 moles P₄O₆

In mass (Molar mass: 220g/mol):

0.0265 moles P₄O₆ * (220g / mol) =

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

hope this mark me brainliest

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l