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3 years ago

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At stp how many liters of oxygen are required to react completely with 3.6 liters

1 answer:

Alex787 [66]3 years ago

3 0

If it is assumed that this reaction is taking place at STP (standard temperature and pressure) then one mole of gas occupies 22.4L

<span>You can then calculate that 3.6L of Hydrogen contains 0.16mol </span>
<span>1mol/22.4L x 3.6L = 0.16mol </span>

<span>Every 2 moles of Hydrogen require 1 mole of oxygen to form water. Therefore half of the number of moles of oxygen are required. </span>
<span>0.16mol/2 = 0.08mol </span>

<span>Again, we know that at STP 1 mol occupies 22.4L </span>
<span>0.08mol x 22.4L/mol = 1.79L of O2 needed. </span>

<span>Also 3.6/2 = 1.8 All ideal gasses will occupy the same ammount of space at STP</span>

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Q2. 0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht

Vera_Pavlovna [14]

Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

Explanation:

No. of moles = $\frac{mass}{molar mass}$

= $\frac{0.254}{204}$ = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= $\frac{0.001245}{0.02}$ = 0.062 mol/L

We know that:

$C_{1} V_{1}$ = $C_{2} V_{2}$

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>

<u><em>= 2.49g/L.</em></u>

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