Answer:
5.52 g
Explanation:
First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 6.30 g NH₃ ÷ 17 g/mol = 0.370 mol NH₃
- 1.80 g O₂ ÷ 32 g/mol = 0.056 mol O₂
Now we <u>calculate with how many NH₃ moles would 0.056 O₂ moles react</u>, using the<em> stoichiometric coefficients</em>:
- 0.056 mol O₂ *
= 0.045 mol NH₃
As there more NH₃ moles than required, NH₃ is the excess reactant.
Then we calculate how many NH₃ moles remained without reacting:
- 0.370 mol NH₃ - 0.045 mol NH₃ = 0.325 mol NH₃
Finally we convert NH₃ moles into grams:
- 0.325 mol NH₃ * 17 g/mol = 5.52 g
Answer:
37
Explanation:
( 98.6 - 32 ) × 5(100c) ÷ 9(180f) = 37
I believe the best answer to that question wud be D. I cud b wrong
Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
