Question

4 pts

Answer 1

First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:

$\frac{1mole}{69.723g} *\frac{43.9g}{1} =0.63MOLga$

So we are dealing with 0.63 moles of gallium metal.

We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:

$\frac{4molesGa}{3moles02} =\frac{0.63molesga}{xmoleso2}$

Cross multiply and solve for x:

4x=1.89

x=47 molesO₂

So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.