Answer:
4Cr + 3O2 —> 2Cr2O3
Explanation:
Information from the question include:
Chromium + oxygen -> chromium(III) oxide
From the word equation given above, the equation can be written as follow:
Cr + O2 —> Cr2O3
The equation can be balance by doing the following:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:
Cr + 3O2 —> 2Cr2O3
Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:
4Cr + 3O2 —> 2Cr2O3
Now the equation is balanced
Answer:
lead ii nitrate is the answer
Answer:
177.1 L
Explanation:
The excersise can be solved, by the Ideal Gases Law.
P . V = n . R . T
In first step we need to determine the moles of gas:
We convert T° from, C° to K → 20°C + 273 = 293K
We convert P from mmHg to atm → 760 mmHg = 1atm
1Dm³ = 1L → 190L
We replace: 190 L . 1 atm = n . 0.082 . 293K
(190L.atm) / 0.082 . 293K = 7.91 moles.
We replace equation at STP conditions (1 atm and 273K)
V = (n . R .T) / P
V = (7.91 mol . 0.082 . 273K) / 1atm = 177.1 L
We can also make a rule of three:
At STP conditions 1 mol of gas occupies 22.4L
Then, 7.91 moles will be contained at (7.91 . 22.4) /1 = 177.1L
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
I am pretty sure it is 4.002602
