Consider the balanced reaction: 4ga(s) + 3o2(g)⟶ 2ga2o3(s) how many moles of oxygen gas will completely react with 43.9 g of gallium metal?

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:

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:

Cross multiply and solve for x:

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

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