How many grams of NH3 are required to make 52.7g H2O in the following reaction : 4NH3+6NO->5N2+6H2O
The stoichiometry problems are worked the same way. Just like yur NG problem.
Step 1. Write and balance the equation.
Step 2. Convert grams of what you have to mols. mols = grams/molar mass if solid or M x L = mols if liquid.
Step 3. Convert mols of what you have to mols of what you want.
Step 4. Convert mols of what you want (from step 3) to grams. g = mols x molar mass.
To find out how many grams of NH3 are required to make 52.7g of H2O in the given reaction (4NH3 + 6NO -> 5N2 + 6H2O), we need to use stoichiometry. Stoichiometry is a way of determining the ratios of substances in a chemical reaction.
Step 1: Determine the molar mass of the given substances.
- The molar mass of NH3 (ammonia) is calculated by adding up the atomic masses of nitrogen (N) and hydrogen (H):
Molar mass of NH3 = (1 mol N) + (3 mol H)
= (14.01 g/mol) + (3 x 1.01 g/mol)
= 17.03 g/mol
- The molar mass of H2O (water) is calculated by adding up the atomic masses of hydrogen (H) and oxygen (O):
Molar mass of H2O = (2 mol H) + (1 mol O)
= (2 x 1.01 g/mol) + (16.00 g/mol)
= 18.02 g/mol
Step 2: Determine the ratio of NH3 to H2O based on the balanced equation.
From the reaction equation, we see that 4 moles of NH3 form 6 moles of H2O.
So, the ratio of NH3 to H2O is 4:6, which simplifies to 2:3.
Step 3: Calculate the grams of NH3 required to produce 52.7g of H2O.
Using the ratio from step 2 (2:3), we can set up a proportion:
(2 g NH3) / (3 g H2O) = (x g NH3) / (52.7 g H2O)
Cross multiply and solve for x:
2 * 52.7 g NH3 = 3 * x g H2O
105.4 g NH3 = 3x
x = 105.4 g NH3 / 3
x ≈ 35.1 g NH3
Therefore, approximately 35.1 grams of NH3 are required to produce 52.7 grams of H2O in the given reaction.