Chemistry, 22.01.2021 18:30 hma9153

If 25 g of AlBr3 was produced whenCsBr was added to the reaction vessel, what quantity (grams) of CsBr was added?

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Answer from: nolangriffin

The question is incomplete, the complete question is;

3CsBr + Al(OH)3 → AlBr3 + 3CsOH

If 25 g of AlBr3 was produced when 125 mL of a CsBr solution was added, what was the molarity of the CsBr solution?

2.25 M

Explanation:

Number of moles of AlBr3 produced = 25g/266.69 g/mol = 0.0937 moles of AlBr3 .

Now, from the reaction equation;

3 moles of CsBr yields 1 mole of AlBr3

Therefore, x moles of CsBr will yield 0.0937 moles of AlBr3

x = 3 * 0.0937 = 0.2811 moles

If 125 mL of a CsBr solution was added

And;

n =CV

where;

n = number of moles

C = concentration of solution

V = volume of solution

C = n/V

C = 0.2811 moles * 1000/125

C= 2.25 M

Answer from: Quest

Answer; carbon monoxide (co);

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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If 25 g of AlBr3 was produced whenCsBr was added to the reaction vessel, what quantity (grams) of Cs...