The moles of I₂ will form from the decomposition of 3.58g of NI₃ is 0.0136 moles.
<h3>How we calculate moles?</h3>
Moles of any substance will be calculated as:
n = W/M, where
W = required mass
M = molar mass
Given chemical reaction is:
2NI₃ → N₂ + 3I₂
Moles of 3.58g of NI₃ will be calculated as:
n = 3.58g / 394. 71 g/mol = 0.009 moles
From the stoichiometry of the solution, it is clear that:
2 moles of NI₃ = produce 3 moles of I₂
0.009 moles of NI₃ = produce 3/2×0.009=0.0136 moles of I₂
Hence, option (3) is correct i.e. 0.0136 moles.
To know more about moles, visit the below link:
brainly.com/question/15303663
Complete Question:
A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.
Answer:
13 mol/L
Explanation:
The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:
M = n/V
The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:
n = 134/207.319
n = 0.646 mol
So, for a volume of 50 mL (0.05 L), the concentration is:
M = 0.646/0.05
M = 12.92 mol/L
Rounded to 2 significant digits, M = 13 mol/L
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
