For the reaction 2 C4H10 + 13 O2 >>>>> 8 CO2 + 10 H2O, if 59.60 g of CO2 is produced, how much C4H10 was needed? Answer should include 2 decimal places.

Answers

Answer 1

19.673g Mass of C4H10 needed to produce 59.60 g of CO2

2C4H10 + 13O2 → 8CO2 + 10H2O

From this Balanced chemical equation we get that ;

2 mol C4H10 produce = 8 mol CO2

By using this relation we can solve given problem .

Mass of CO2 = 51.60 g

Moles of CO2 = Mass of CO2 / Molar mass of CO2

= 59.60 g / 44.01 (g/mol)

= 1.354 moles

now:

2 mol C4H10 produce = 8 mol CO2

So, X mol C4H10 produce = 1.354 mol CO2

X mol C4H10 =(2mol C4H10×1.354 mol)/8 mol CO2

= 0.3385 mol CO2

Moles of C4H10 =0.3385 mol

Mass of C4H10= Moles of C4H10 × Molar mass of C4H10

=0.3385mol × 58.12 (g/mol)

= 19.673 g

19.673g Mass of C4H10 needed

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