URGENT** If you have 35.0g of Na and 100.0g of Cl2, what is the maximum amount of NaCl that you can produce?

2Na + Cl2 → 2NaCl

89.0 g

165 g

75.8g

57.7 g

165 g of NaCl are formed in the reaction

Explanation:

2Na + Cl₂ → NaCl

In order to determine the limiting reactant, we convert the mass of each reactant to moles

35 g / 23g/mol = 1.52 moles Na

100 g / 70.9 g/mol = 1.41 moles Cl₂

1 mol of chlorine reacts with 2 moles of Na, so If I have an x value of moles of Cl₂ I would need the double to react.

For 1.41 moles of Cl₂, I need 2.82 moles of Na; therefore my limiting reagent is the Na. Ratio is 2:2. So if I have 2.82 moles of Na I will produce 2.82 moles of NaCl

We convert the moles to mass: 2.82 mol . 58.45 g/1 mol =164.8 g

We can produce 89.0 grams of NaCl

Explanation:

Step 1: Data given

Mass of Na = 35.0 grams

Mass of Cl2 = 100.0 grams

Molar mass of Na = 22.99 g/mol

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Na + Cl2 → 2NaCl

Step 3: Calculate moles Na

Moles Na = mass Na / molar mass Na

Moles Na = 35.0 grams / 22.99 g/mol

Moles Na = 1.52 moles

Step 4: Calculate moles Cl2

Moles Cl2 = mass Cl2 / molar mass Cl2

Moles Cl2 = 100.0 grams / 70.9 g/mol

Moles Cl2 = 1.41 moles

Step 5: Calculate limiting reactant

For 2 moles Na we need 1 mol Cl2 to produce 2 moles NaCl

Na is the limiting reactant. It will completely be consumed ( 1.52 moles).

Cl2 is in excess. There will react 1.52/2 = 0.76 moles

There will remain 1.41 - 0.76 = 0.65 moles

Step 6: Calculate moles of NaCl

For 2 moles Na we need 1 mol Cl2 to produce 2 moles NaCl

For 1.52 moles Na we'll have 1.52 moles NaCl

Step 7: Calculate mass NaCl

Mass NaCl = 1.52 moles* 58.44 g/mol

Mass NaCl =  88.8 grams ≈ 89 grams

We can produce 89.0 grams of NaCl

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years