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Vaselesa [24]
3 years ago
14

Consider the balanced reaction: 4Ga(s) + 3O2(g)⟶ 2Ga2O3(s)

Chemistry
1 answer:
Paha777 [63]3 years ago
5 0

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}=0.63mol_{Ga}


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{4moles_{Ga}}{3moles_{O_{2}}} =\frac{0.63moles_{Ga}}{xmoles_{O_{2}}}


Cross multiply and solve for x:


4x=1.89


x=0.47moles_{O{2}}


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

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