Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO
/ Kg C
H
- 15.3848 Kg air / kg C H
Explanation:
Molar masses of each product are :
Butane = 58 kg /kmol
Oxygen = 32 kg/kmol
Nitrogen = 28 kg/kmol
water = 18 kg/kmol
<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>
= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )
= 176 / 950.32
= 0.1852
<em><u>2) Calculate the mass fraction of water </u></em>
= ( 5 * 18 ) / (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))
= 90 / 950.32
= 0.0947
<em><u>3) Calculate the mass fraction of Nitrogen </u></em>
= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))
= 684.32 / 950.32
= 0.7201
<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>
Mco2 = ( 4 * 44 ) / 58 = 3.0345 kg CO / Kg C H
<u>5) Mass of Air required per unit of fuel mass burned </u>
Mair = ( 6.5 * 32 + 24.44 *28 ) / 58 = 15.3848 Kg air / kg C H
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
Answer:
Explanation:
A marble is not a very large object, so a smaller graduated cylinder is a better choice. A 100 milliliter graduated cylinder is not needed to measure the volume of a small marble, so a 25 milliliter graduated cylinder is the best option.
2 meters per second. You do 10 divided by 5 to find your answer, which is 2.
If a hypothesis is stated and outcome of the experiment is what was predicted, then it supports the hypothesis. if the experiment does not support the hypothesis, then the outcome was not what was predicted.
