Suppose 50.0 g NH3, 80.0 g CO2, and 2.00 mol H2O are reacted. Determine the

maximum grams of NH4HCO3 that can be produced.

This is a limiting reagent (LR) problem and you know that because amounts are given for all of the reactants instead of just one.

NH3 + CO2 + H2O ==> NH4HCO3

I do these the long way.
1. How much product can we get from 50.0 g NH3 if all of the others are in excess?
mols NH3 = g/molar mass = estimated 3 mol but this is an estimate and you should rdo this and all of the calculations that follow more accurately.
Then 3 mols NH3 x (1 mol NH4HCO3/1 mol NH3) = 3 x 1/1 = 3 mols NH4HCO3.

2. How much product can we get from 80.0 g CO2 if all of the others are in excess?
mols CO2 = g/molar mass = estimated 1.8 mols CO2.
Then 2 mols CO2 x (1 mol NH4HCO3/1 mol CO2) = 1.8 x 1/1 = estimated 1.8 mols NH4HCO3.

3. How much product can we get from 2.00 mls H2O if all of the others are in excess?
2.00 mols H2O x (1 mol NH4HCO3/1 mol H2O) = 2.00 x 1/1 = 2.00 mols NH4HCO3.

You can see that the values for NH4HCO3 vary so only one of those can be right. The correct value in LR problems is ALWAYS the smallest number and the reagent responsible for that number is the LR. So the LR is CO2 and estimated 1.8 mols NH4HCO3 will be formed.

Convert that to grams. grams = mols x molar mass = ?

To determine the maximum grams of NH4HCO3 that can be produced, we need to determine which reactant is limiting and use its stoichiometry with NH4HCO3 to calculate the maximum yield.

1. Calculate the number of moles of each reactant:
- NH3:
Given Mass of NH3 = 50.0 g
Molar mass NH3 = 17.03 g/mol
Number of moles of NH3 = (50.0 g NH3) / (17.03 g/mol) = 2.93 mol NH3

- CO2:
Given Mass of CO2 = 80.0 g
Molar mass CO2 = 44.01 g/mol
Number of moles of CO2 = (80.0 g CO2) / (44.01 g/mol) = 1.82 mol CO2

- H2O:
Given Number of moles of H2O = 2.00 mol H2O

2. Write the balanced chemical equation for the reaction:
NH3 + CO2 + H2O -> NH4HCO3

3. Determine the limiting reactant:
To determine the limiting reactant, compare the ratios of the reactants and their coefficients in the balanced equation. The reactant with the smallest mole ratio compared to the coefficient ratio is the limiting reactant.

Ratio for NH3: 1 mol NH3 : 1 mol NH4HCO3
Ratio for CO2: 1 mol CO2 : 1 mol NH4HCO3

Since NH3 has a greater mole ratio-to-coefficient ratio than CO2, NH3 is not the limiting reactant.

Ratio for H2O: 1 mol H2O : 1 mol NH4HCO3

Since H2O has a smaller mole ratio-to-coefficient ratio than both NH3 and CO2, H2O is the limiting reactant.

4. Calculate the maximum moles of NH4HCO3 formed:
From the balanced equation, we know that the ratio of H2O to NH4HCO3 is 1:1. Therefore, the number of moles of NH4HCO3 formed will be equal to the number of moles of H2O.

Maximum moles of NH4HCO3 = Number of moles of H2O = 2.00 mol

5. Calculate the maximum grams of NH4HCO3 formed:
Molar mass of NH4HCO3 = 79.06 g/mol
Maximum grams of NH4HCO3 = (2.00 mol) * (79.06 g/mol) = 158.12 g

Therefore, the maximum grams of NH4HCO3 that can be produced is 158.12 g.

To determine the maximum grams of NH4HCO3 that can be produced, we need to find the limiting reactant in the given reaction. The limiting reactant is the one that is completely consumed and limits the amount of product that can be formed.

Let's start by writing the balanced equation for the reaction:
NH3 + CO2 + H2O → NH4HCO3

From the balanced equation, we can see that the molar ratio between NH3 and NH4HCO3 is 1:1. This means that for every 1 mole of NH3, we can produce 1 mole of NH4HCO3.

Given:
Mass of NH3 = 50.0 g
Mass of CO2 = 80.0 g
Moles of H2O = 2.00 mol

To find the limiting reactant, we need to convert the masses of NH3 and CO2 to moles.

Molar mass of NH3 = 17.03 g/mol
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3

Number of moles of NH3 = 50.0 g / 17.03 g/mol ≈ 2.93 mol (rounded to two decimal places)

Molar mass of CO2 = 44.01 g/mol
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2

Number of moles of CO2 = 80.0 g / 44.01 g/mol ≈ 1.82 mol (rounded to two decimal places)

Now, we compare the moles of NH3 and CO2 to the moles of H2O.

Mole ratio of NH3 to H2O in the equation is 1:1, which means that for every 1 mole of NH3, we need 1 mole of H2O.

Since we have 2.00 moles of H2O, and both NH3 and CO2 have more moles than H2O, we can conclude that the moles of H2O do not limit the reaction.

Next, we compare the moles of NH3 and CO2.

The mole ratio of NH3 to CO2 in the equation is also 1:1, which means that for every 1 mole of NH3, we need 1 mole of CO2.

Since we have more moles of NH3 (2.93 mol) than CO2 (1.82 mol), the CO2 is the limiting reactant in this reaction.

To find the maximum grams of NH4HCO3 that can be produced, we need to calculate the moles of NH4HCO3 produced from the limiting reactant.

From the balanced equation, we can see that the molar ratio between CO2 and NH4HCO3 is 1:1. This means that for every 1 mole of CO2, we can produce 1 mole of NH4HCO3.

Since we have 1.82 moles of CO2 (the limiting reactant), the moles of NH4HCO3 produced will also be 1.82 mol.

Finally, we need to convert the moles of NH4HCO3 to grams by using the molar mass of NH4HCO3.

Molar mass of NH4HCO3 = 79.06 g/mol
Mass of NH4HCO3 = Moles of NH4HCO3 × Molar mass of NH4HCO3

Mass of NH4HCO3 = 1.82 mol × 79.06 g/mol ≈ 144.02 g (rounded to two decimal places)

Therefore, the maximum grams of NH4HCO3 that can be produced in this reaction is approximately 144.02 grams.