Chemistry, 06.06.2020 05:00 kamy8900

How many liters of H2 at STP are required to react with 2.3 g of Fe3O4 to give FeO and H2O?

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Answer from: ashah1260

0.22L of H2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Fe3O4 + H2 —> 3FeO + H2O

Next, we shall determine the number of mole present in 2.3g of Fe3O4. This is illustrated below:

Molar mass of Fe3O4 = (56x3) + (16x4) = 232g/mol

Mass of Fe3O4 = 2.3g

Mole of Fe3O4 =..?

Mole = Mass/Molar Mass

Mole of Fe3O4 = 2.3/232 = 9.91x10^-3 mole.

Next, we shall determine the number of mole H2 needed for the reaction. This is illustrated below :

From the balanced equation above,

1 mole of Fe3O4 reacted with 1 mole of H2.

Therefore, 9.91x10^-3 mole of Fe2O3 will also react with 9.91x10^-3 mole of H2.

Finally, we can obtain the volume of H2 required for the reaction as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, 9.91x10^-3 mole of H2 will occupy = 9.91x10^-3 x 22.4 = 0.22L

Therefore, 0.22L of H2 is required for the reaction.

Answer from: Quest

answer: option (4) is the correct answer.

explanation:

areas which are at low altitudes and near the equator tend to receive direct sunlight from the sun. hence, heat energy or radiations from the sun make these areas warmer and they have warmest water.

whereas areas which are at high altitudes and away from the equator do no receive direct sunlight. hence, no direct contact from the sun makes these regions colder and thus, they have colder water.

thus, we can conclude that the statement areas at low latitudes near the equator have the warmest water because they receive the most direct sunlight best explains the relationship between an area's geography and the temperature of its surface water.

Answer from: Quest

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

= 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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