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ra1l [238]
3 years ago
11

2A + 3B -->C + D When 5 units of A and 6 units of B are allowed to react, the limiting reactant would be:

Chemistry
1 answer:
kodGreya [7K]3 years ago
7 0

Answer:

The limiting reactant is B.

Explanation:

The equation for the reaction is given below:

2A + 3B —> C + D

The limiting reactant can be obtained as follow:

From the equation above,

2 units of A reacted with 3 units of B.

Therefore, 5 units of A will react with = (5 x 3) /2 = 7.5 units of B.

From the calculation made above, we can see that it will take a higher unit of B i.e 7.5 units than what was given i.e 6 units to react completely with 5 units of A.

Therefore, B is the limiting reactant and A is the excess reactant.

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3 years ago
Determine the total pressure of a mixture of 0.400 mole of He and 0.600 mole of Ne in a 2.00 liter container at 25oC.
harina [27]

Answer:

Total pressure of the mixture is 12.2 atm

Explanation:

Let's apply the Ideal Gases law to solve this

Total pressure . V = Total moles . R . T

Total moles = 0.4 m of He and 0.6 mole of Ne → 1 mol

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8 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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4 years ago
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