A major source of boron is the mineral kernite, Na2[ B4 O5(OH)4] ·3 H2 O . Calculate how many gr

step 1 So a major source of boron is the mineral kernite. Kernite is an A2, B4, 05, oh H4 times 3H2O. O

A major source of boron is the mineral kernite, Na2[ B4...

A major source of boron is the mineral kernite, $\mathrm{Na}_{2}\left[\mathrm{~B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O} .$ Calculate how many grams of boron can be produced from $1.0 \times 10^{3} \mathrm{~kg}$ of a kernite-bearing ore if the ore contains $0.98 \%$ kernite by mass and the process has a $65 \%$ yield.

A major source of boron is the mineral kernite, $\mathrm{Na}_{2}\left[\mathrm{~B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O} .$ Calculate how many grams of boron can
be produced from $1.0 \times 10^{3} \mathrm{~kg}$ of a kernite-bearing ore if the ore contains $0.98 \%$ kernite by mass and the process has a $65 \%$ yield.

Key Concepts

Percent Composition

Percent composition refers to the fraction of a material’s mass that is made up of a specific component. In geochemical and material extraction contexts, understanding the percent composition of a valuable mineral within an ore is essential to determine how much of the desired element can ultimately be obtained.

Molar Mass and Mass-to-Mole Conversion

Molar mass is a critical concept in chemistry that connects the mass of a substance to the number of moles. Converting the mass of a mineral to moles using its molar mass allows for stoichiometric calculations which are necessary when determining how many atoms or molecules of an element are present in the mineral.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It is used to determine the yield of a desired product based on the balanced chemical equation and the composition of the starting material.

Reaction Yield

Reaction yield represents the efficiency of a chemical process, expressed as a percentage. It accounts for the practical losses and incomplete reactions by comparing the actual amount of product obtained to the theoretical maximum amount, making it a crucial factor in scaling up laboratory procedures to industrial production.

Transcript

00:02 So a major source of boron is the mineral kernite.

00:06 Kernite is an a2, b4, 05, oh h4 times 3h2o.

00:24 Okay, calculate how many grams of boron can be produced from 1 .0 times 10 to do its power of 3 kilograms of kernite bearing ore.

00:32 If the ore contains 0 .98 % curnite bimath and the process has a 65 % yield.

00:39 So we have a 0 .98 % kernite by mass and we have a 65%.

00:58 So we're asked to find the grams of boron that are produced.

01:09 Okay.

01:12 So we stated that there are 1 .0 times 10 feet of power 3 kilograms of, kernite in our, or not kernite, yeah, of kernite bearing ore.

01:29 So let's start with this.

01:31 So we want to find the percent, or not the percent, the grams of bond.

01:36 So what we need to first do is multiply this gram or kilogram of burnite bearing ore with 0 .98 % kernite by mass to get the amount of kernite in our ore.

01:51 So 1 .0 times 10 to the power of 3 times .98 % gives me 9 .8 kilograms of kernin.

02:13 Now we wanted our percent, or not our percent, are willing to find the grams of boron.

02:21 So right now we have it in kilograms, so let's convert that to grams.

02:26 So what we need to do is multiply 9 .8 kilograms.

02:32 Times 10 to the power of 3, oh sorry, 10 to the power of 3 grams over 1 kilogram.

02:41 So that gives me 9 .8 divided by 10 to the power of 3, which gives me 9 .8 times 10 to the power of 1, 2, 3, 0 ,000, 0 ,000 of cronite.

03:01 Sorry about that, actually.

03:03 It could be the other way around because we're trying to find grams...

Please give Ace some feedback