Module 4.3 Equation Calculations

Question 1

Equation Calculation

C₄H₁₀ + O₂———> CO₂ + H₂O

2.9g of C₄H₁₀———> ?g CO₂

Answer 1

First Balance the equation
C₄H₁₀ + 6 1/2 O₂———> 4CO₂ + 5H₂O
Multiply all coefficients by 2 (the denominator for 6 1/2)
2C₄H₁₀ + 13O₂———> 8CO₂ + 10H₂O

Then use periodic table to find weight for each element * subscript 
C₄H₁₀ = 12*4=48 + 10*1=10 = 58mw
O₂ = 16*2= 32mw
CO₂ = 12*1= 12 + 16*2= 32 = 44mw
H₂O = 1*2= 2 + 1*16= 16 = 18mw

Then use this to answer a problem
ex. 2.9g of C₄H₁₀———> ?g CO₂

Use the given info (2.9g) and convert to moles using the mw we found
2.9g /58mw = 0.05mol for C₄H₁₀

Using this new mol (0.05mol) find the moles for the ?g CO₂

Use the coefficient from the known substance 2C₄H₁₀ as the denominator and the coefficient from the unknown substance as the numerator 8CO₂
⁸/₂

Take the mole 0.05 * 8 / 2 = 0.20mol

Now convert from mol to g to solve problem

0.20 * 44mw = 8.8g

Answer= 2.9g of C₄H₁₀———> 8.8g CO₂