If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium phosphate will produced? How many grams of excess reactant will be left over?

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

First we convert the mass of both reactants to moles, using their respective molar masses:

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

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We calculate the produced moles of Na₃PO₄ using the limiting reactant:

Then we convert moles into grams:

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We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

We substract that amount from the original amount:

Finally we convert those remaining moles to grams:

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