# Eli Ingram, Geneva Mathews, Gracie Schindler
# 10/24/2023
# Project 3


# Problem 3.8A
# a*(C2H6) + b*(O2) --> c*(CO2) + d*(H2O)     This is the unbalanced equation.
var ('a b c d')

# This equation balances the number of carbon atoms on both sides of the reaction.
2*a == c

# Problem 3.8B
# These equations balance the number of hydrogen and oxygen atoms in the reaction. 
6*a == 2*d

2*b == 2*c + d

# Problem 3.8C
Rxn = matrix([[2,0,1,0],[6,0,0,2],[0,2,2,1]])
print(Rxn.rref()*6)
# There are infinitely-many solutions because there is a free-variable present. 
# We multiplied this coefficeint matrix by 6 because we cannot have a fraction molecule in our reaction.

# Solution to this system: a = -1/3(d), b = -7/6(d), c = 2/3(d)
# 2(C2H6) + 7(O2) --> 4(CO2) + 6(H2O)     This is the balanced equation.




# Problem 3.9A
# a*(C6H12O6) + b*(O2) --> c*(CO2) + d*(H2O)    This is the unbalanced equation.
var ('a, b, c, d')

# Set up equations that balances the number of carbon atoms on both sides of the reaction.
6*a == c

2*b == 2*c + d

c == 6*a

2*d == 12*a

Rxn = matrix([[6,0,-1,0],[12,0,0,-2],[6,2,-2,-1]])
print(Rxn.rref()*6)

# By balancing the carbon and hydrogen atoms, and thus the equation we get this balanced chemical equation:
# 1(C6H12O6) + 6(O2) --> 6(CO2) + 6(H2O)




# Problem 3.9B
# a*(Mg) + b*(CO2) --> c*(MgO) + d*(C)    This is the unbalanced equation.
var ('a, b, c, d')

a == c

b == d

2*b == c

Rxn = matrix([[0,1,0,1],[1,0,1,0],[0,2,1,0]])
print(Rxn.rref())