The volume of N₂ at STP=56 L
<h3>Further explanation</h3>
Given
2.5 moles of N₂
Required
The volume of the gas
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, the volume per mole of gas or the molar volume-Vm is 22.4 liters/mol.
So for 2.5 moles gas :
The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams
First, find moles of gold given the mass of the sample:
(35.9g Au)/(197.0g/mol Au) = 0.182mol Au
Second, multiply moles of Au by Avogrado's number:
(0.182mol)(6.02 x10^23)= 1.10x10^23 atoms Au
I'm not so sure but I would say Answer Choice B
