6

我有一个大学作业，我必须平衡以下等式：

NaOH + H2S04 --> Na2S04 + H20

目前，我对 python 和一般编码的了解非常有限。到目前为止，我已经尝试使用矩阵来求解方程。看起来我得到了解决方案 a=b=x=y=0 我想我需要将其中一个变量设置为 1 并解决其他三个。我不知道该怎么做，我进行了搜索，看起来其他人使用了更复杂的代码，而我真的无法遵循它！

这是我到目前为止所拥有的

    #aNaOH + bH2S04 --> xNa2SO4 +y H20

    #Na: a=2x
    #O: a+4b=4x+y
    #H: a+2h = 2y
    #S: b = x

    #a+0b -2x+0y = 0
    #a+4b-4x-y=0
    #a+2b+0x-2y=0
    #0a +b-x+0y=0

    A=array([[1,0,-2,0],

             [1,4,-4,-1],

             [1,2,0,-2],

             [0,1,-1,0]])

    b=array([0,0,0,0])




    c =linalg.solve(A,b)

    print c

0.0.0.0
4

4 回答 4

4

问题是您已经构建了一个线性系统，其中b是一个零向量。现在对于这样的系统，总是有一个直截了当的答案，即所有变量也都是零。由于将一个数字与零相乘并将零相加总是会产生零。

一种解决方案可能是将 1 分配给变量。举个例子a。如果我们分配a = 1，那么我们将得到b，x并且y函数为a1。

所以现在或线性系统是：

 B  X  Y |    #
    2    |1   #  A    = 2X
-4  4  1 |1   #  A+4B = 4X+4Y
-2     2 |1   #  A+2B =    2Y
-1  1  0 |0   #     B =     X

或将其放入代码中：

>>> A = array([[0,2,0],[-4,4,1],[-2,0,2],[-1,1,0]])
>>> B = array([1,1,1,0])
>>> linalg.lstsq(A,B)
(array([ 0.5,  0.5,  1. ]), 6.9333477997940491e-33, 3, array([ 6.32979642,  2.5028631 ,  0.81814033]))

这意味着：

 A = 1, B = 0.5, X = 0.5, Y = 1.

如果我们将其乘以 2，我们得到：

2 NaOH + H2S04 -> Na2S04 + 2 H20

哪个是对的。

于 2017-07-20T16:25:14.823 回答
1

我在 Python 中提到了Solve system of linear integer equations，它被翻译成

# Find minimum integer coefficients for a chemical reaction like
#   A * NaOH + B * H2SO4 -> C * Na2SO4 + D * H20
import sympy
import re

# match a single element and optional count, like Na2
ELEMENT_CLAUSE = re.compile("([A-Z][a-z]?)([0-9]*)")

def parse_compound(compound):
    """
    Given a chemical compound like Na2SO4,
    return a dict of element counts like {"Na":2, "S":1, "O":4}
    """
    assert "(" not in compound, "This parser doesn't grok subclauses"
    return {el: (int(num) if num else 1) for el, num in ELEMENT_CLAUSE.findall(compound)}

def main():
    print("\nPlease enter left-hand list of compounds, separated by spaces:")
    lhs_strings = input().split()
    lhs_compounds = [parse_compound(compound) for compound in lhs_strings]

    print("\nPlease enter right-hand list of compounds, separated by spaces:")
    rhs_strings = input().split()
    rhs_compounds = [parse_compound(compound) for compound in rhs_strings]

    # Get canonical list of elements
    els = sorted(set().union(*lhs_compounds, *rhs_compounds))
    els_index = dict(zip(els, range(len(els))))

    # Build matrix to solve
    w = len(lhs_compounds) + len(rhs_compounds)
    h = len(els)
    A = [[0] * w for _ in range(h)]
    # load with element coefficients
    for col, compound in enumerate(lhs_compounds):
        for el, num in compound.items():
            row = els_index[el]
            A[row][col] = num
    for col, compound in enumerate(rhs_compounds, len(lhs_compounds)):
        for el, num in compound.items():
            row = els_index[el]
            A[row][col] = -num   # invert coefficients for RHS

    # Solve using Sympy for absolute-precision math
    A = sympy.Matrix(A)    
    # find first basis vector == primary solution
    coeffs = A.nullspace()[0]    
    # find least common denominator, multiply through to convert to integer solution
    coeffs *= sympy.lcm([term.q for term in coeffs])

    # Display result
    lhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(lhs_strings)])
    rhs = " + ".join(["{} {}".format(coeffs[i], s) for i, s in enumerate(rhs_strings, len(lhs_strings))])
    print("\nBalanced solution:")
    print("{} -> {}".format(lhs, rhs))

if __name__ == "__main__":
    main()

运行起来像

Please enter left-hand list of compounds, separated by spaces:
NaOH H2SO4

Please enter right-hand list of compounds, separated by spaces:
Na2SO4 H2O

Balanced solution:
2 NaOH + 1 H2SO4 -> 1 Na2SO4 + 2 H2O
于 2017-07-20T19:13:55.780 回答
0

您可以使用此解决方案。它适用于任何化学方程式。最后一个系数可以用 b[i]!=0 的行计算

H2SO4+NaOH−−>Na2SO4+H2OH2SO4+NaOH−−>Na2SO4+H2O

a=np.array([[2,1,0],[1,0,-1],[4,1,-4],[0,1,-2]])
b=np.array([2,0,1,0])
x=np.linalg.lstsq(a,b,rcond=None)[0]
print(x)

y=sum(x*a[0])/b[0]   
print("y=%f"%y)

出去：

[0.5 1. 0.5] y=1.000000

于 2018-08-31T15:47:45.290 回答
-1

做得太好了。然而，当我在 David Lay 的线性代数教科书第 5 版中对以下方程进行测试时，我收到了一个可以进一步简化的次优解决方案。

在页。55, 1.6 练习检查 ex 7.:

NaHCO_3 + H_3C_6H_5O_7 --> Na_3C_6H_5O_7 + H_2O + CO_2

您的代码段返回：

平衡解决方案：

15NaHCO3 + 6H3C6H5O7 -> 5Na3C6H5O7 + 10H2O + 21CO2

正确答案是：

3NaHCO_3 + H_3C_6H_5O_7 -> Na_3C_6H_5O_7 + 3H_2O + 3CO_2
于 2018-05-21T00:23:26.590 回答