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2 years ago

In which of the following is the solution concentration expressed in percent by volume

Chemistry

2 answers:

Hitman42 [59]2 years ago

8 0

Answer : The correct option is, $\frac{10g\text{ of solute}}{100ml\text{ of solution}}\times 100\%$

Explanation :

The solution concentration expressed in percent by volume means that the amount of solute present in 100 parts volume of solution.

It is represented in formula as :

$\text{Percent solution by volume}=\frac{\text{Amount of solute}}{\text{Amount of solution}}\times 100\%$

Or,

$\text{Percent solution by volume}=\frac{10g\text{ of solute}}{100ml\text{ of solution}}\times 100\%$

Hence, the correct option is, $\frac{10g\text{ of solute}}{100ml\text{ of solution}}\times 100\%$

sleet_krkn [62]2 years ago

4 0

10ml of solute/100ml of solution *100% third one

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In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

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3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$