ammonia (NH3) chemically reacts with oxygen gas (O2) to produce nitric oxide (NO) and water (H2O). What mass of water is produced by the reaction of 1.09g of oxygen gas?
Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
2NH3+5/2O2--->2NO+3H2O
5/2X32 3X18
80 54
1.09 ?
[APPROX 1]
CROSS MUTIPPLY IF U TAKE IT AS 1.09
1X54/80 THEN,1.09X54/80=0.73575
=54/80
=27/40
=0.675
First, write a balanced chemical equation for the reaction:
→+4NH35O2+4NO6H2O
.
The mole ratio of any two compounds in a chemical reaction is given by the ratio of the stoichiometric coefficients of those compounds in the balanced chemical equation. That means this equation is true:
moles of
H2O
produced
moles of
NH3
consumed
=
6
4
.
Knowing this key equation, use the following plan to answer the question.
Convert grams of ammonia consumed to moles using the molar mass of ammonia.
Calculate moles of water produced from moles of ammonia consumed using the mole ratio above.
Convert moles of water produced to grams using the molar mass of water.
Step 1: Find the moles of ammonia consumed.
You'll need the molar mass of ammonia:
1
×
molar mass of
N
+
3
×
molar mass of
H
1
molar mass of
NH3
1
×
14.0067·gmol−1
+
3
×
1.00794·gmol−1
17.03052·gmol−1
Note: Be sure you use the atomic masses from the ALEKS
Calculator or ALEKS Periodic Table and do not round early so
that your answer has the correct number of significant digits.
Use the molar mass to convert grams of ammonia consumed to moles of ammonia produced:
moles of
NH3
consumed
=
(mass of
NH3
consumed)
(molar mass of
NH3
)
=
9.05g17.03052·gmol−1
=
0.53140…mol
.
There are
3
significant digits in
9.05g
. That means there are
3
significant digits in your calculated answer for moles of ammonia. However, keep a few extra digits for now and only round your final answer.
Step 2: Find the moles of water produced.
Multiply the moles of ammonia consumed by the mole ratio:
moles of
H2O
produced
=
(moles of
NH3
consumed)
×
(moles of
H2O
produced)
(moles of
NH3
consumed)
=
×0.53140mol64
=
0.79710…mol
.
Because stoichiometric coefficients are exact numbers, the number of moles of water has the same number of significant digits as the number of moles of ammonia.
Step 3: Find the mass of water produced.
You'll need the molar mass of water:
2
×
molar mass of
H
+
1
×
molar mass of
O
1
molar mass of
H2O
2
×
1.00794·gmol−1
+
1
×
15.9994·gmol−1
18.01528·gmol−1
Use the molar mass to convert moles of water produced to grams of water produced:
mass of
H2O
produced
=
(moles of
H2O
produced)
×
(molar mass of
H2O
)
=
×0.79710mol18.01528·gmol−1
=
14.35998…g
.
There are
3
significant digits in your calculated answer for moles of water. That means there are
3
significant digits in your final calculated answer.
Be sure to round your final answer correctly to
3
significant digits.
To find the mass of water produced in the reaction, we need to calculate the moles of oxygen gas and then determine the mole ratio between oxygen gas and water in the balanced chemical equation.
Let's begin by calculating the number of moles of oxygen gas (O2) using its molar mass. The molar mass of O2 is 32 g/mol (16 g/mol for each oxygen atom).
To find the number of moles, we divide the mass of oxygen gas by its molar mass:
Number of moles of O2 = Mass of O2 / Molar mass of O2
Number of moles of O2 = 1.09 g / 32 g/mol ≈ 0.03406 mol (rounded to five decimal places)
Now let's determine the mole ratio between oxygen gas (O2) and water (H2O) using the balanced equation:
4NH3 + 5O2 → 4NO + 6H2O
From the balanced equation, we see that the mole ratio between O2 and H2O is 5:6. This means that for every 5 moles of O2, we get 6 moles of H2O.
Using this mole ratio, we can calculate the number of moles of water produced:
Number of moles of H2O = Number of moles of O2 × (6 moles H2O / 5 moles O2)
Number of moles of H2O = 0.03406 mol × (6/5) ≈ 0.04087 mol (rounded to five decimal places)
Finally, we need to convert the moles of water to its mass using the molar mass of water, which is 18 g/mol (2 g/mol for each hydrogen atom and 16 g/mol for oxygen).
Mass of H2O = Number of moles of H2O × Molar mass of H2O
Mass of H2O = 0.04087 mol × 18 g/mol ≈ 0.736 g (rounded to three decimal places)
Therefore, the mass of water produced by the reaction of 1.09 g of oxygen gas is approximately 0.736 grams.