7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:
1 answer:
Answer:
4.03dm³
Explanation:
The reaction expression is given as:
3H₂ + N₂ → 2NH₃
Volume of hydrogen = 12dm³
AT rtp:
1 mole of gas occupies volume of 22.4dm³
x mole of hydrogen will occupy a volume of 12dm³
Number of moles of hydrogen = = 0.54mole
From the balanced reaction equation:
3 mole of hydrogen gas combines with 1 mole of Nitrogen gas
0.54 mole of hydrogen as will therefore combine with = 0.18moles of nitrogen gas
Since ;
1 mole of gas occupies a volume of 22.4dm³
0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³
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Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.
Explanation:
According to the dilution law,
where,
= concentration of stock solution = 12.0 M
= volume of stock solution = ?
= concentration of diluted solution= 6.00 M
= volume of diluted acid solution = 500 ml
Putting in the values we get:
Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.