From the stoichiometry of the reaction, carbon is in excess and 5.856 g s left over.
<h3>What is the volume of butane produced?</h3>
The reaction can be written as; 4C(s) + 5H2(g) -----> C4H10(g)
Number of moles of C = 13.45 g/1 2g/mol = 1.12 moles
If 1 mole of hydrogen occupies 22.4 L
x moles of hydrogen occupies 17.65 L
x = 0.79 moles
Now;
4 moles of carbon reacts with 5 moles of hydrogen
1.12 moles of carbon reacts with 1.12 moles * 5 moles/4 moles
= 1.4 moles of hydrogen
Hence hydrogen is the limiting reactant here and carbon is in excess.
If 4 moles of carbon reacts with 5 moles of hydrogen
x moles of carbon reacts with 0.79 moles of hydrogen
x = 0.632 moles
Number of moles of carbon unreacted = 1.12 moles - 0.632 moles
= 0.488 moles
Mass of carbon unreacted = 0.488 moles * 12 g/mol
= 5.856 g
Volume of butane produced is obtained from;
5 moles of hydrogen produces 1 mole of butane
0.79 moles of hydrogen produces 0.79 moles * 1 mole/ 5 moles
= 0.158 moles
1 mole of butane occupies 22.4 L
0.158 moles of butane occupies 0.158 moles * 22.4 L/ 1 mole
= 3.53 L
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