Question

given the following balanced chemical equation how many grams of aluminum chloride are produced when 3.00 moles of aluminum reacts with 5.3 moles of chlorine gas. Al+Cl2->AlCl3

Answer 1

Consider putting coefficients in front of each compound:\[
xAl+yCl_3 \to zAlCl3
\]We basically want to solve for \(x\), \(y\), and \(z\).
There are some equations we can derive from this formula: \[
\begin{array}{rcl}
x &=& z \\
2y &=& 3z
\end{array}
\]The problem here though is that we only have 2 equations.
In fact even when we have more equations that coefficients, some of the equations will be redundant and we will be one equation short.

The reason for this is because there are actually multiple solutions to balancing this equation. If we found a solution \(x_1\), \(y_1\), and \(z_1\), then multiplying them all by \(2\) would also be a solution.

So we want the smallest solution in which all coefficients are real numbers.

Answer 2

What well do is tack on some variable \(n\), and solve for all the variables in terms of \(n\).
Then we can chose an \(n\) that fits best. \[
\begin{array}{rcl}
n &=& x &&&& x &=& n \\
x &=& z &&&& z &=& n\\
2y &=& 3z & 2y &=& 3n & y &=& \frac{3}{2}n
\end{array}
\]We need to set \(n = 2\) so that \(y\) doesn't end up being a fraction. Doing this gives us: \[
\begin{array}{rcl}
x &=& (2) &=& 2 \\
z &=& (2) &=& 2 \\
y &=& \frac{3}{2}(2) &=& 3
\end{array}
\]

Answer 3

Putting this back into our equation gets us:\[
xAl+yCl_3 \to zAlCl3 \\
2Al+3Cl_3 \to 2AlCl3
\]We now have \(2Al\) and \(6Cl\) on both sides. Our equation is balanced.