How many grams of NO gas can be produced from 80.0mL of 4.00M HNO3 and excess Cu?
3Cu + 8HNO3---3Cu(NO3)2 + 4H20 + NO
This is a stoichiometry problem. Here is an example of a stoichiometry problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the amount of NO gas produced, we need to calculate the number of moles of HNO3 and use the stoichiometry of the balanced chemical equation to determine the number of moles of NO. Finally, we can convert moles of NO to grams.
Step 1: Calculate the number of moles of HNO3.
To do this, we need to use the given volume and molarity of HNO3.
Molarity (M) = moles (mol) / volume (L)
We need to convert mL to L, so divide the volume by 1000.
80.0 mL / 1000 = 0.080 L
Using the formula, we can find moles:
Moles of HNO3 = Molarity x Volume
Moles of HNO3 = 4.00 M x 0.080 L
Moles of HNO3 = 0.320 mol
Step 2: Use stoichiometry to find moles of NO.
From the balanced chemical equation, we see that the stoichiometric ratio of HNO3 to NO is 8:1.
This means that for every 8 moles of HNO3, we get 1 mole of NO.
Moles of NO = Moles of HNO3 / Ratio of HNO3 to NO
Moles of NO = 0.320 mol / 8
Moles of NO = 0.040 mol
Step 3: Convert moles of NO to grams.
To convert moles to grams, we need to use the molar mass of NO, which is 30.0061 g/mol.
Mass of NO = Moles of NO * Molar mass of NO
Mass of NO = 0.040 mol * 30.0061 g/mol
Mass of NO = 1.200 g
Therefore, 1.200 grams of NO gas can be produced from 80.0 mL of 4.00M HNO3 and excess Cu.