3 Cu + 8HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O

Question

3 Cu  +   8HNO₃ → 3 Cu(NO₃)₂   +  2 NO  +  4  H₂O

In the above equation, how many grams of water can be made when 14.3 moles of HNO₃ are consumed?

 

Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element	Molar Mass
Hydrogen	1
Nitrogen	14
Copper	63.5
Oxygen	16

Answer 1

3 Cu + 8HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O

In the above equation, how many grams of water can be made when 14.3 moles of HNO3 are consumed?

Round your answer to the nearest tenth

given: 14.3 mol HNO3

desired: grams of water:

strategy:

moles HNO3 → mol H2O → grams H2O

So, there's two steps:

Step 1:

14.3 moles HNO3 → ? mol H2O

Use the coefficients from the balanced equation:

8 mol HNO3 = 4 mol H2O

Therefore:

(^{14.3 mol HNO3} / ₁ ) (^{4mol H2O} / _{8 mol HNO3})=7.15 mol H2O

Step 2:

7.15 mol H2O → ? g H2O

Use the molar mass of H2O

the molar mass of H2O depends ONLY on the formula, and DOES NOT depend on the coefficients:

molar mass H2O=2(H)+O

2(1.01)+16.00=18.02 g/mol H2O, so

1 mol H2O=18.02 g H2O

Therefore:

(^{7.15 mol H2O} / ₁ ) (^{18.02 g H2O} / _{1 mol H2O})=128.843 g H2O

(OR , use the molar mass equation)

mol = grams / molar mass

therefore

grams = mol (molar mass)

grams = 7.15mol(18.02 g/mol)=128.843 g H2O

round to the tenths:

128.843 g H2O

128.8 g H2O

Answer 2

Since you're determining the mass of water, first you must determine its molar mass: H₂O = 18.0 g/mol.

Next, you use the balanced equation to 'compare' the given info to the desired.

(18.0 g H₂O/1 mol H₂O) (4 mol H₂O/8 mol HNO₃) 14.3 mol HNO₃ = 128.7 g H₂O