what mass of water will be formed by reaction of 3.50g of methylamine (8CH3NH2) with 15.0g of perchloric acid (13HClO4)? Answer is 99.1. How do you work this?
*answer is 4.14g
See your post above.
To find the mass of water formed in this reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely consumed, limiting the amount of product formed.
First, let's write the balanced chemical equation for the reaction between methylamine (CH3NH2) and perchloric acid (HClO4):
8CH3NH2 + 13HClO4 -> 11H2O + 8(CH3NH3ClO4)
From the balanced equation, we can see that the mole ratio between methylamine and water is 8:11. We will use this information to determine the number of moles of water formed.
Step 1: Calculate the number of moles of methylamine:
Molar mass of methylamine (CH3NH2) = (12.01 g/mol * 1) + (1.01 g/mol * 4) + (14.01 g/mol * 1) = 31.06 g/mol
Number of moles of methylamine (CH3NH2) = mass / molar mass = 3.50 g / 31.06 g/mol = 0.1127 mol
Step 2: Calculate the number of moles of perchloric acid:
Molar mass of perchloric acid (HClO4) = (1.01 g/mol * 1) + (35.45 g/mol * 1) + (16.00 g/mol * 4) = 100.46 g/mol
Number of moles of perchloric acid (HClO4) = mass / molar mass = 15.0 g / 100.46 g/mol = 0.1494 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the moles of reactants using their respective coefficients from the balanced equation. The reactant with fewer moles compared to the stoichiometric ratio will be the limiting reactant.
Moles of water (H2O) formed from methylamine = (0.1127 mol methylamine) * (11 mol water/8 mol methylamine) = 0.1558 mol
Moles of water (H2O) formed from perchloric acid = (0.1494 mol perchloric acid) * (11 mol water/13 mol perchloric acid) = 0.1261 mol
Since the mole ratio between methylamine and water is 8:11, the moles of water formed from methylamine (0.1558 mol) are greater than the moles of water formed from perchloric acid (0.1261 mol). Therefore, perchloric acid is the limiting reactant.
Step 4: Calculate the mass of water formed:
Molar mass of water (H2O) = (1.01 g/mol * 2) + (16.00 g/mol) = 18.02 g/mol
Mass of water formed = moles of water * molar mass of water = 0.1261 mol * 18.02 g/mol = 2.272 g
Therefore, the mass of water formed by the reaction of 3.50 g of methylamine with 15.0 g of perchloric acid is 2.272 g.
(Note: The given answer of 99.1 g seems to be incorrect based on the calculations.)