Chemistry, 06.10.2019 04:30 christianfoster17

One of the intermediates in the synthesis of glycine from ammonia , carbondioxide and methane is aminoacetonitrile c2h4n2. the balanced chemical equation is 3ch4+5 co2 + 8nh3 > 4c2h4n2 +10h2o. how much c2h4n2 could be expected from the reaction of 13.2 g co2, 2.18 g nh3 and 17.0 g ch4

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Answer from: marika35

mass of C₂H₄N₂ = 3.472 g

Explanation:

We have the following chemical reaction:

3 CH₄ + 5 CO₂ + 8 NH₃ → 4 C₂H₄N₂ + 10 H₂O

Using the masses given by the problem we calculate the number of moles for each reactant:

number of moles = mass / molecular weight

number of moles of CO₂ = 13.2 / 44 = 0.3 moles

number of moles of NH₃ = 2.18 / 17 = 0.13 moles

number of moles of CH₄ = 17 / 16 = 1.06 moles

We can see that the limiting reactant is ammonia NH₃. Now we can devise the following reasoning:

if 8 moles of NH₃ produces 4 moles of C₂H₄N₂

then 0.13 moles of NH₃ produces X moles of C₂H₄N₂

X = (0.13 × 4) / 8 = 0.062 moles of C₂H₄N₂

mass of C₂H₄N₂ = number of moles × molecular weight

mass of C₂H₄N₂ = 0.062 × 56 = 3.472 g

Answer from: Quest

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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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