N2H4+2H2O2->N2+4H2O
a) If you start with 23.0 g of hydrazine and 23.0g of hydrogen peroxide, which substance is the limiting reactant?
b) How much of the other chemical is left unreacted?
c) Assuming 75% yield, how many grams of water are formed?
I have no idea how to do this problem. I can't remember anything chemistry and need someone to walk me through it please.
Correction: Under part a, that is supposed to be 25.0 grams hydrogen peroxide.
You need to compute how many moles of each you have. Then, the balanced equation tells you that you need twice the moles of peroxide. Do you have more than twice? if so, you have an excess of proxide. If you have less than twice the moles, you have a deficit of peroxide.
If you then use the deficit (limiting) reactant as the starting number of moles, you know how many moles of product you should have gotten. However, multipy that by .75 as it was not efficient conversion.
Sure! I'd be happy to help you with this problem.
To determine the limiting reactant and find the amounts of substances involved in the reaction, we will follow these steps:
Step 1: Write the balanced chemical equation.
Step 2: Calculate the number of moles for each reactant.
Step 3: Use the mole ratios from the balanced equation to determine the limiting reactant.
Step 4: Calculate the amount of unreacted chemical (excess reactant) for the limiting reactant.
Step 5: Calculate the amount of product formed using the limiting reactant and the percent yield.
Let's break it down step by step.
Step 1: Write the balanced chemical equation.
The balanced chemical equation given is:
N2H4 + 2H2O2 -> N2 + 4H2O
Step 2: Calculate the number of moles for each reactant.
To do this, we need the molar masses of the substances involved.
The molar mass of hydrazine (N2H4) = 32.05 g/mol
The molar mass of hydrogen peroxide (H2O2) = 34.02 g/mol
Using these molar masses, we can convert the given masses into moles:
For hydrazine:
Number of moles = mass / molar mass = 23.0 g / 32.05 g/mol
For hydrogen peroxide:
Number of moles = mass / molar mass = 23.0 g / 34.02 g/mol
Step 3: Use the mole ratios from the balanced equation to determine the limiting reactant.
Looking at the balanced equation, we can see that the mole ratio between hydrazine and hydrogen peroxide is 1:2. This means that for every mole of hydrazine, we need two moles of hydrogen peroxide.
So, we compare the moles of each reactant to see which one is limiting.
Step 4: Calculate the amount of unreacted chemical (excess reactant) for the limiting reactant.
To do this, we need to find the amount of the excess reactant that is not consumed in the reaction. We can use the mole ratio given in the balanced equation to calculate this.
Step 5: Calculate the amount of product formed using the limiting reactant and the percent yield.
First, we need to determine the limiting reactant from Step 3. The reactant that produces fewer moles of product is the limiting reactant.
Once we know the limiting reactant, we can use the mole ratio from the balanced equation to determine the moles of product formed. Finally, we can convert the moles of product to grams using the molar mass of water (H2O).
We need the percent yield to calculate the actual amount of water formed. The percent yield tells us the percentage of the theoretical yield (amount of product calculated from the balanced equation) that is actually obtained in the experiment. In this case, the percent yield is given as 75%.
Using these steps, we can answer all the specific questions:
a) If you start with 23.0 g of hydrazine and 23.0 g of hydrogen peroxide, which substance is the limiting reactant?
To answer this, we need to compare the moles of hydrazine and hydrogen peroxide calculated in Step 2. The limiting reactant is the one that produces fewer moles of product.
b) How much of the other chemical is left unreacted?
To answer this, we use the mole ratio from the balanced equation to determine the excess reactant.
c) Assuming 75% yield, how many grams of water are formed?
To determine the mass of water formed, we use the mole ratio from the balanced equation to calculate the moles of water, and then convert it to grams using the molar mass of water. Finally, we multiply by the percent yield to obtain the actual amount of water formed.
Let me know if you would like me to walk you through the calculations for each step!