"You have 17.981g of AgNO3 reacts with 15.70g of water vapour and 12.71g of nitrogen monoxide"
3AgNO3(aq) + 2H2O(g) + NO(g) --> 3Ag(s) + 4HNO3(aq)
First you need to identify the limiting reagent (LR). I do these the long way. I've estimated and rounded so you need to check everything including molar masses and my estimates.
1. mols AgNO3 = g/molar mass = 17.981/169.9 = 0.105. How many mols HNO3 is produced if AgNO3 is the LR?
That's 0.105 mols AgNO3 x (4 mols HNO3/3 mols AgNO3) = 0.105 x 4/3 = 0.14
2. mols H2O = 15.70/18 = 0.872
How many mols HNO3 produced if H2O is the LR? That's
0.872 x (4 mols HNO3/2 mols H2O) = 1.74
3. mols NO = 12.71/30 = 0.424
mols HNO3 produced if NO is the LR. That's
0.424 x (4 mols HNO3/1 mol NO) = 1.70
You check all of this but if I've not made a mistake (it's past my bed time) I see 0.14 mol HNO3 for AgNO3; 1.74 for H2O; 1.70 for NO. The smallest number is 0.14 for AgNO3 so that's the LR. Grams HNO3 produced is mols HNO3 x molar mass HNO3 = 0.14 x 63 = ?
Please help! I tried to answer the question below and couldn't figure out the steps I needed to follow:
When 17.981g of AgNO3 reacts with 15.70g of water vapour and 12.71g of nitrogen monoxide, solid silver and nitric acid are produced. Answer the following questions based on the given chemical equation below: 3AgNO3(aq) + 2H2O(g) + NO(g) --> 3Ag(s) + 4HNO3(aq)
a) Identify the limiting reactant based on the above chemical reaction
b)Find the mass of the nitric acid, HNO3, produced based on the above chemical reaction
3 answers
how about the mass of the product produced?
Did you read anything I wrote? The product you want is HNO3. Read that last sentence. It is "Grams HNO3 produced is mols HNO3 x molar mass HNO3 = 0.14 x 63 = ?"
The only thing I didn't do was pick up a calculator and multiply 0.14 x 63. I thought you could do that.
The only thing I didn't do was pick up a calculator and multiply 0.14 x 63. I thought you could do that.