"\\begin{aligned}\nCaCO_3 + HNO_3 \\to Ca(NO_3)_2 + H_2O + CO_2\n\\end{aligned}"
To balance the chemical reaction using algebraic method,
"\\begin{aligned}\naCaCO_3 + bHNO_3 \\to cCa(NO_3)_2 + dH_2O + eCO_2\n\\end{aligned}"
for Calcium --- a = c
for Carbon --- a = e
for Oxygen --- 3a + 3b = 6c + d + 2e
for Hydrogen --- b = 2d
for Nitrogen --- b = 2c
To solve, we have to set one of the coefficients to a fixed number.
Let us set d = 1
for Hydrogen;
b = 2d
b = 2(1)
"\\therefore" b = 2
for Nitrogen;
b = 2c
2 = 2c
"\\therefore" c = 1
for calcium;
a = c
"\\therefore" a = 1
for carbon;
e = a
"\\therefore" e = 1
Let's put all coefficients into Oxygen and find out if our coefficients are actually correct
"\\therefore" For Oxygen;
3a + 3b = 6c + d + 2e
3(1) + 3(2) = 6(1) + 1 + 2(1)
3 + 6 = 6 + 1 + 2
9 = 9
"\\therefore" a = 1, b = 2, c = 1, d = 1, e = 1
and the balanced chemical reaction is;
"\\begin{aligned}\nCaCO_3 + 2HNO_3 \\to Ca(NO_3)_2 + H_2O + CO_2\n\\end{aligned}"
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