What is the limiting reactant when 19.9g of CuO are exposed to 2.02g of H2 according to the following equation?
CuO + H2 ---> Cu + H20
My attempt at this:
19.9gCuO * 1moleCuO/79.539gCuO * 1moleCu/1moleCuO = .250moleCu
2.02gH2 * 1moleH2/2.016gH2 * 1moleCu/1moleH2 = 1.00moleCu
The first one is limiting reactant?
B. How many grams of Cu are produced?
.250moleCu * 63.54gCu/1moleCu = 15.9gCu
Is that correct?
Both A and B look ok to me.
Thanks!
Well, if you ask me, analyzing stoichiometry can sometimes feel like trying to solve a complex puzzle. But fear not, my friend, I'm here to add a little laughter to the mix!
Based on your calculations, it seems like you've got your answers right. So, congratulations, you're on the right track!
But let's just hope the reactants don't throw a mini-celebration party before the reaction takes place. After all, we don't want the CuO and H2 to say, "Hey, let's limit ourselves and not give Clown Bot a chance to make silly jokes!"
So remember, when it comes to limiting reactants, it's all about finding the one that's feeling a bit shy, hiding in the corner, and holding back the reaction. Good luck with your chemistry, and enjoy the comedic journey!
Yes, your calculations are correct.
Based on the stoichiometry of the reaction, you have determined that 19.9 g of CuO corresponds to 0.250 moles of CuO, and 2.02 g of H2 corresponds to 1.00 mole of H2.
Since the stoichiometric ratio between CuO and H2 is 1:1, the reactant that produces fewer moles of Cu is the limiting reactant. In this case, CuO produces 0.250 moles of Cu, while H2 produces 1.00 mole of Cu. Therefore, CuO is the limiting reactant.
To determine the mass of Cu produced, you can use the stoichiometric ratio between Cu and CuO, which is 1:1. From the previous calculation, we found that 0.250 moles of Cu is produced.
0.250 moles of Cu * 63.54 g/mole = 15.9 g of Cu
Therefore, 15.9 grams of Cu are produced in this reaction.
Your attempt is almost correct, just a small mistake in the calculation. Let's go through it step by step:
To determine the limiting reactant, we need to compare the number of moles of Cu produced from each reactant.
1. Calculate the number of moles of Cu produced from CuO:
19.9g CuO * (1 mol CuO/79.539g CuO) * (1 mol Cu/1 mol CuO) = 0.250 mol Cu
2. Calculate the number of moles of Cu produced from H2:
2.02g H2 * (1 mol H2/2.016g H2) * (1 mol Cu/1 mol H2) = 0.999 mol Cu
Since CuO produces 0.250 moles of Cu and H2 produces 0.999 moles of Cu, we can see that CuO is the limiting reactant because it produces fewer moles of Cu.
Now, let's calculate the mass of Cu produced:
0.250 mol Cu * (63.54g Cu/1 mol Cu) = 15.9g Cu
So, the correct answer is 15.9g of Cu produced.