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olchik [2.2K]
3 years ago
11

What is the mass of water necessary to generate 11.2 L of hydrogen gas if calcium metal reacts with water at standard temperatur

e and pressure (STP)?
Chemistry
1 answer:
12345 [234]3 years ago
5 0

Answer:

The mass of water is 18 g

Explanation:

The reaction of calcium with water can be represented in the equation below:

Ca + 2H₂O --------->Ca(OH)₂ +H₂

1 Mole of gas at STP = 22.4L

From the displacement reaction above, calculate the mass of water that will produce 22.4L of hydrogen gas at STP.

Mass of 2H₂O = 2(2x1 + 16) = 2X18 = 36 g/mol

Using proportional analysis;

36 g of 2H₂O produced 22.4 L of H₂, then

what mass of 2H₂O will produce 11.2L of  H₂ ?

Mathematically,

22.4 L ----------------------------------> 36g

11.2 L -----------------------------------> ?

Cross and multiply, to obtain the expression below

= (11.2 X 36)/22.4

= 18 g

Therefore, the mass of water is 18 g

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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
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0.00370 g

Explanation:

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To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

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\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

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3 0
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