Respuesta :

nozevairby777
 (a) The balanced equation shows that 2 moles of water result from 5 moles of 
oxygen that reacts, so that is 0.4 times as many moles of water as O2. So if we 
have 2.50 moles of oxygen reacting, we produce 0.4 times that amount, or 
1.0 mole of water. 
(b) The molecular weight of acetylene is 26 and that of O2 is 32, so we can set 
up the proportion: 2.25 gm. / 52 = x gm. / 160 
Then x grams = (2.25)(160) / 52 = 6.923 grams of O2 are required. 
(c) The balanced equation shows that twice as many moles of CO2 are produced 
as we have acetylene reacting. If 78.0 grams of acetylene react, that is 3 moles 
so we produce twice that, or 6 moles of CO2. The molecular weight of CO2 is 44, 
so we have 44 times 6, or 264 grams of carbon dioxide produced. 
(d) If we collect 186 grams of CO2, the percentage yield is: 
186/264 = 0.7045, or 70.45 percent yield. 
Whew! Hope this answers all parts of your question!
kenmyna

 The  grams   of oxygen that  is required  to react   completely    with  859.0 g C₂H₂  is 2643.2 grams


calculation

 Step 1: write the equation for reaction   between C₂H₂  and O₂

that is 2C₂H₂ +5O₂  →   4 CO₂ +2H₂O

Step 2 : find the moles of C₂H₂

moles =  mass/molar mass

From  periodic table the  molar mass  of C₂H₂  = (12 x2) +( 1 x2)  = 26 g/mol

moles= 859.0 g /26 g/mol =33.04 moles


step 3 : use the  moles ratio  to determine the  moles of  O₂

C₂H₂:O₂  is    2:5  therefore the  moles  of O₂ =33.04  x5/2 = 82.6 moles


Step 4: find   mass  of O2

 mass =  moles x  molar mass

from periodic table the  molar mass of O₂ = 16 x2 = 32 g/mol

mass = 82.6  moles x 32 g/mol =2643.2 grams

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