2H2O(l) + N2H4(l) -----> 4H2O(g) + N2(g) determine how many grams of N2 are produced from the reaction of 8.37g of H2O2 and 5.74g of N2H4.

Answer 1

2H₂O₂(l) + N₂H₄(l) -----> 4H₂O(g) + N₂(g)

Step 1) Make sure you have a balanced chemical reaction equation. It's a good thing I checked this because you had a typo in your equation (the initial H₂O₂ was written as H₂O).

Step 2) Calculate molar masses:

N₂ = 14.007 x 2 = 28.014 g/mol

H₂O₂ = 1,008 x 2 + 15.999 x 2 = 34.014 g/mol

N2H4 = 14.007 x 2 + 1.008 x 4 = 32.046 g/mol

Step 3) Calculate moles available of the compounds in question:

8.37g of H₂O₂: 8.37/34.014 = 0.24608 moles

5.74g of N₂H₄: 5.74/32.046 = 0.17912 moles

Step 4) Using molar ratios in the chemical reaction equation, calculate moles of the compound in question:

There are 2 moles of H₂O₂ required for each 1 mole of N₂H₄ and so the 0.24608 moles of H₂O₂ is the limiting reactant (only 0.12304 moles of N₂H₄ are required so there is excess N₂H₄).

For every 2 moles of H₂O₂, 1 mole of N₂ is produced, so, since there are 0.24608 moles of H₂O₂ then 0.12304 moles of N₂ is produced.

Step 5) Convert moles to grams:

0.12304 moles of N₂: 0.12304(28.014) = 3.45 grams

Answer 2

2H₂O₂(l) + N₂H₄(l) -----> 4H₂O(g) + N₂(g)

Since we are given the amounts of BOTH reactants, this becomes a limiting reactant problem. We must first find which reactant, if either, is limiting. An easy way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For H₂O₂: 8.37 g H₂O₂ x 1 mol H₂O₂ / 34 g = 0.246 moles H₂O₂(÷2->0.123)

For N₂H₄: 5.74 g N₂H₄ x 1 mol N₂H₄ / 32 g = 0.179 moles N₂H₄(÷1->0.179)

Since 0.123 is less than 0.179, H₂O₂ is the limiting reactant.

Grams N₂ produced = 0.246 moles H₂O₂ x 1 mole N₂ / 2 mols H₂O₂ x 28 g N₂/mol = 3.44 g N₂