1. With how many grams of hydrogen did you have to start the reaction if you wan

Question

1. With how many grams of hydrogen did you have to start the reaction if you want to produce 50.0g of product?
N2 (g) + 3 H2 (g) -> 2 NH3 (g)

2. How many grams of product would you get if you start with 30.0 g of nitrogen and 20.0 g of hydrogen?

N2 (g) + 3 H2 (g) -> 2 NH3 (g)

3. If your experiment actually produced 40.0 g of product starting with 30.0g of nitrogegn, what is the percent yield?
N2 (g) + 3 H2 (g) -> 2 NH3 (g)

4. How many grams of Na2CO3 do you need to react with 50. mL of 1.50 M HCl solution?
Na2CO3 (aq) + 2 HCl (aq) -> 2 NaCl (aq) + H2O (l) + CO2 (g)

5. What is the concentration of 50. mL of HCl that would react with 75 mL of 1.50 M Na2CO3?
Na2CO3 (aq) + 2 HCl (aq) -> 2 NaCl (aq) + H2O (l) + CO2 (g)

I need help with all of the questions but I don't understand how to do question number 5 at all. Need work shown on that one.

Explanation / Answer

For the first part:

To solve these, first, determine the balanced equation: (you've already got that).
Then determine the number of moles of each reactant.
From the moles of provided reacants and the ratios from the balanced equation you should be able to determine which is the limiting reagent (the one that runs out first), Note: sometimes the exact amounts are given, so no excess reactants, but you need to check.
From the moles of limiting reagant and the ratios from the balanced equation you should be able to calculate the number of moles of product that could be produced. (If they give you a % yield for the reaction, make certian to take it into account when calculating moles/grams of final product). I'm not certian where on earth you got those numbers for part one, so I will do that one for you and you can find another in the text book to practice on.

N2 + 3 H2 ---> 2 NH3

You have 28g of N2 so you need to determine moles of N2. Do this by dividing weight of N2 by the molar nass for N2 (Remember N weighs 14 but nitrogen gas is N2 and weighs 28g/mole!)

So the moles are:

For nitrogen, 28g/28g/mol = 1.0 mole N2

For hydrogen 25g/2g/mole = 12.5 moles

(Remember, hydrogen gas is H2 not H; so it weighs 2 x 1g per mole).

Obviously, hydrogen is in excess and nitrogen is the limiting reagant. As, from the ratios in the balanced equation, we can see that 3 moles of hydrogen can react with our 1 mole of nitrogen to give 2 moles of ammonia so all the rest of the hydrogen is excess.

To get moles of ammonia we use the moles of limiting reagant (N2) and the ratio from the balanced equation:

1.0 mole N2 x (2.0 moles NH3/1.0 mole N2) = 2.0 moles NH3

To get grams of final product, just multiply the moles of final product by the molecular weight:

2.0 moles NH3 x [14 + (3 x 1)]g/mole = 2.0 x 17g/mole = 34g

If you need to know the amount of leftover starting material(s) just subtract the moles that reacted from the starting amount:

For H2: 12.5 moles - 3 moles H2 = 9.5 moles H2

Sometimes, they ask for the quantities of "leftovers" in grams instead of moles so just multiply the number of moles by the molar mass of the material. If one of the products is a liquid and they want the volume; to get it from moles, first multiply by the molar mass to get grams, then divide by the density in g/ml to get volume in ml. (Note: 1 ml = 1 cm^3). There are numerous variations on these questions.

For the second part:

This one is done similar to the previous one (you need to balance the equation, determine moles of reactants and the limiting reagant (if any). Then you can determine the moles of the products (including the hydrogen).

So, you now got moles of hydrogen and you want grams there are 2 ways to do this: In this case they specify the conditions for the hydrogen, as "standard conditions" or STP, so we can use the rule that 1 mole of any gas occupies 22.4 liters at STP. Remember this; it will save you extra work on a lot of your calculations! Remember the 22.4 liters only is true when the gas is at STP! To get the volume of the gas just take the moles produced and multiply by 22.4 liters/mole; the mole units cancel and you get the volume in liters.

If the gas is at conditions other than STP they will generally give the temperature and pressure (or some means to determine same) and you will need to use the ideal gas law; PV = nRT where you solve for volume: V = nRT/P. Then plug in the known values. Make certian you use the correct gas constant (usually, you want the R value that is 0.08206Latm/molK) and have P & T in correct units! Then you will get your volume in liters.

Some variations of these equations they may give you volume (or even mass) of gas produced under some conditions of T & P and want you to calculate amount of starting material before reaction; or they may give that, and ask for percent yield of the reaction etc. Many many variations!

In some situations, such as the last one, when it is relatively simple, you don't even have to convert to moles just use the ratios of the weights product/starting material.

2 KClO3 ---> 2 KCl + 3 O2

KClO3 weighs 39.09 + 35.45 + (3 x 16.00) = 122.54g/mole
KCl weighs 39.09 + 35.45 = 74.54g/mole

25g x (74.54/122.54) = 15.21 g KCl

If they wanted the weight of oxygen produced it would be:

25g x [(3 x 16)/122.54] = 9.79g O2

You can also do these by using the grams to moles, then moles back to grams method as well. It would be best if you did these that way untill you are familiar with which situations you can use the "easy way" and which times you can't.

Using this method you get:

74.54 g/mole x (2/2) x (25g/122.54g/mole) = 15.21g

A reaction usually is 2 or more materials "reacting" to form 1 or more products. A decomposition is where a single starting material is broken down into 2 or more products. At least thats my opinion. In terms of solving the problems, it really doesn't make much difference.

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