If 120 grams of aluminum oxide reacts with 100 grams of magnesium, which reactant is in excess and by how much? Al2O3 + 3Mg —> 3MgO + 2Al

Limiting reactant is magnesium.

14.1 g remain unreacted.

Explanation:

Hello!

In this case, given the balanced chemical reaction, we are able to compute the moles of each reactant by using their molar masses:

Now, since they are reacting based on a 1:3 mole ratio, we can compute the moles of magnesium consumed by the 1.18 moles of aluminum oxide as follows:

Meaning there are more available moles of magnesium than consumed, and therefore it is the limiting reactant. Moreover, it is in excess by:

It means that 14.1 g of magnesium remain unreacted.

Regards!

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2