Question

A) What volume of butane (C 4 H 10 ) can be produced at STP, from the reaction of 13.45 g of carbon with 17.65 L of hydrogen gas at STP?
b) Which reactant is in excess and how much of this reactant is left over?

Answer 1

From the stoichiometry of the reaction, carbon is in excess and 5.856 g s left over.

What is the volume of butane produced?

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

Learn more about stoichiometry:brainly.com/question/9743981

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