LPG Burns In the air according to the equation C3H3 + 5O2 —> 3CO2 + 4H2O When 50

Question

LPG Burns In the air according to the equation
C3H3 + 5O2 —> 3CO2 + 4H2O
When 500.0 g of LPG was used to burn in the air 1000.g of carbon dioxide was produced what was the percent yield of that reaction?
500lpgx 1 mol /44 x 3molCO2/1 x 44.01/1
100/1500g x 100 = 66.7%
Is there an easier way to set this up ??
A short cut maybe ? LPG Burns In the air according to the equation
C3H3 + 5O2 —> 3CO2 + 4H2O
When 500.0 g of LPG was used to burn in the air 1000.g of carbon dioxide was produced what was the percent yield of that reaction?
500lpgx 1 mol /44 x 3molCO2/1 x 44.01/1
100/1500g x 100 = 66.7%
Is there an easier way to set this up ??
A short cut maybe ?
C3H3 + 5O2 —> 3CO2 + 4H2O
When 500.0 g of LPG was used to burn in the air 1000.g of carbon dioxide was produced what was the percent yield of that reaction?
500lpgx 1 mol /44 x 3molCO2/1 x 44.01/1
100/1500g x 100 = 66.7%
Is there an easier way to set this up ??
A short cut maybe ?

Explanation / Answer

Answer:

Given balanced equation C3H3+5O2----->3CO2+4H2O

First find the moles of C3H3, given mass of C3H3=500 g and molar mass=39 g/mol

So mol of C3H3=mass/molar mass=500 g/39 g/mol=12.82 mols C3H3

Now from the equation mol ratio between C3H3 and CO2 is 1:3

Then moles of CO2=3xmoles of C3H3=3x12.82 mol=38.46 mol CO2.

Now multiply with molar mass of CO2 we will gwt theoretical yield.

mass of CO2 produced=38.46 molx44 g/mol=1692.30 g

Given experimental yield of CO2=1000 g.

So % yield= (experimental yield/theoretical yield)x100

% yield=(1000 g/1692.3 g)100=59.09 %.

Shortcut is (500g/39g)x(3 mol CO2)x44 g/mol=1692.3 g

% yield=(1000/1692.3)x100=59.09%.

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