Using proportions from chemical equations 🧮
French version 🇫🇷
Complete combustion of butane 🔥
Word equation:
Butane + Dioxygen → Carbon dioxide + Water
Chemical equation:
2
C
4
H
10
+
13
O
2
→
8
CO
2
+
10
H
2
O
Question 1
70 butane molecules are burnt.
How many dioxygen molecules are
needed?
C
4
H
10
O
2
Chemical equation
Calculation:
Result:
Conclusion:
?
dioxygen molecules are needed.
Question 2
70 butane molecules are burnt.
How many water molecules H
2
O
appear?
C
4
H
10
H
2
O
Chemical equation
Calculation:
Result:
Conclusion:
?
water molecules H
2
O appear.
2
13
70
?
13×70/2
13/2×70
70×13/2
70/2×13
13*70/2
13/2*70
70*13/2
70/2*13
\( \mathrm { \displaystyle\frac{70 \times 13}{2} } \)
455
2
10
70
?
10×70/2
10/2×70
70/2×10
70×10/2
10*70/2
10/2*70
70/2*10
70*10/2
\( \mathsf { \displaystyle\frac{70 \times 10}{2} } \)
350
📊 RESULT:
Next exercise 🔀