How many grams of nitrogen monoxide gas can be produced from 80.0 ml of 4.00 M HNO3 and excess Cu?
3Cu(s)+ 8Hno3(aq)-> 3Cu(NO3)2(aq)+4H2O(/)+2NO(g)
Here is a worked example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Can't view the sight can u help me work it out
To find the number of grams of nitrogen monoxide (NO) gas produced, we'll follow these steps:
Step 1: Determine the number of moles of HNO3
We are given the volume (80.0 mL) and the molarity (4.00 M) of HNO3. We can use the formula:
moles = volume (L) × molarity
First, let's convert the volume from milliliters (mL) to liters (L):
80.0 mL = 80.0 ÷ 1000 = 0.080 L
Now, we can calculate the moles of HNO3:
moles of HNO3 = 0.080 L × 4.00 M = 0.32 moles
Step 2: Determine the limiting reactant
To find the limiting reactant, we need to compare the moles of HNO3 to the stoichiometric ratio between HNO3 and NO. From the balanced equation, we know that each mole of HNO3 produces 2 moles of NO.
moles of NO = 0.32 moles × (2 moles NO / 8 moles HNO3) = 0.08 moles
Step 3: Calculate the grams of NO
To convert moles to grams, we need to use the molar mass of NO. The molar mass of NO is:
N: 14.01 g/mol
O: 16.00 g/mol
Molar mass of NO = 14.01 g/mol + 16.00 g/mol = 30.01 g/mol
Now, we can calculate the grams of NO:
grams of NO = moles of NO × molar mass of NO
grams of NO = 0.08 moles × 30.01 g/mol = 2.401 grams
Therefore, approximately 2.4 grams of nitrogen monoxide gas can be produced from 80.0 mL of 4.00 M HNO3 and excess Cu.