If 55 g of SiO2(s) glass is placed into 111 L of 0.032 M HF, is there enough HF to dissolve al

step 1 This question deals with stoichiometry. And any time we have a stoichiometric question, we also

If 55 g of SiO2(s) glass is placed into 111 L of 0.032 M...

If $55 \mathrm{~g}$ of $\mathrm{SiO}_{2}(s)$ glass is placed into $111 \mathrm{~L}$ of $0.032 \mathrm{M} \mathrm{HF}$, is there enough HF to dissolve all of the glass? Determine which substance is the limiting reagent and calculate how much of the other reagent is left if the reaction proceeds to completion.

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to determine the quantitative relationships among reactants and products. This process includes converting measured masses or volumes into moles and then applying the mole ratios from the balanced equation to determine how much of each substance is consumed or produced.

Limiting Reagent

The limiting reagent is the substance that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed. Identifying the limiting reagent is essential for accurate yield calculations, as it limits the extent of the reaction.

Molarity and Volume Relationship

Molarity, defined as moles of solute per liter of solution, is a key concept in solution chemistry. It allows the calculation of the number of moles of a reactant present in a given volume of a solution, which is crucial when dealing with reactions in a liquid medium.

Excess Reagent

The excess reagent is the reactant that remains in surplus after the limiting reagent has been fully consumed. Determining the amount of the excess reagent left over involves subtracting the quantity that reacted (based on the limiting reagent and stoichiometric ratios) from the initial quantity present.

Transcript

00:01 This question deals with stoichiometry.

00:03 And any time we have a stoichiometric question, we also have to start with a balanced reaction.

00:08 So we're told in the question that we have silicon dioxide reacting with hydrofluic acid.

00:14 That's going to produce water in a silicon tetrafluoride product.

00:19 And so first we need to ensure that the left side, all the moles of reactants on the left side, balance properly with the moles of product on the right side of the reaction.

00:30 And they do.

00:31 And so we're starting again with the balanced reaction.

00:36 We're also told in the question that we have 111 liters of hydrofluoric acid that can react.

00:43 And here's the concentration, 0 .32 molar or 0 .032 moles per liter.

00:48 And so we're going to use that balanced reaction at this point to determine how many moles of the silicon tetrafluoride product can be produced from the starting amount first of hydrofluoric acid.

01:04 And the balanced reaction shows us that it requires four moles of hydrofluoric acid to produce one mole of silicon tetrafluoride.

01:12 So let's place that information into all of this information into the balanced reaction.

01:20 And if we crunch the numbers out, we determine that the maximum amount of hydrofluor or silicon tetrafluoride that can be produced by that starting amount of hydrofluoric acid is 0 .89 moles.

01:33 And we're going to do the same thing with the information provided about the other reagent here, the silicon dioxide.

01:40 We're told we have 55 grams of it.

01:42 How many moles is that? let's plug in the molar mass of silicon dioxide.

01:47 It's 60 .09 grams per mole.

01:50 And if we stop the math at this point and looked at our numbers, that would tell us then that we have 0 .92 moles of silicon dioxide available to react in this process.

02:06 And again, just as there is a 4 to 1 molar relationship between the hydrofluoric acid and silicon tetrafluoride, there is a 1 -to -1 relationship...

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