Question

The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2. How many moles of CO2 are produced when 35.0 mol C2H2 react completely?

Answer 1

Answer: 67.2 moles

Explanation: $2C_2H5+5O_2\rightarrow 4CO_2+2H_2O$

According to the given balanced equation, 2 moles of acetylene $C_2H_2$ combine with 5 moles of oxygen $O_2$  to produce 4 moles of carbon dioxide $CO_2$.

Thus if 2 moles of acetylene $C_2H_2$ combine with = 5 moles of oxygen $O_2$

35 moles of acetylene $C_2H_2$ combine with=$\frac{5}{2}\times {35}=87.5$ moles of oxygen $O_2$

But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen $O_2$ reacts with = 2 moles of acetylene $C_2H_2$

84 moles of oxygen $O_2$ reacts with=$\frac{2}{5}\times {84}=33.6$ moles of acetylene $C_2H_2$

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene $C_2H_2$  produce= 4 moles of carbon dioxide $CO_2$.

33.6 moles of acetylene $C_2H_2$ produce=$\frac{4}{2}\times {33.6}=67.2$ moles of of carbon dioxide $CO_2$.

Answer 2

87.5 is the answer

i just took it

the second one is

33.6