Answer:
94.58 g of 
Explanation:
For this question we have to start with the reaction:
Now, we can balance the reaction:
We have the amount of 
and the amount of 
. Therefore we have to find the limiting reactive, for this, we have to follow a few steps.
1) Find the moles of each reactive, using the molar mass of each compound (
).
2) Divide by the coefficient of each compound in the balanced reaction ("2" for and "1" for ).
<u>Find the moles of each reactive</u>
<u>Divide by the coefficient</u>
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The smallest values are for , so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:
We have to remember that the molar ratio between and is 2:2 and the molar mass of is 18 g/mol.
Answer:
d
Explanation:
Firstly, we need to see the theoretical mole ratio between nitrogen and ammonia from the balanced chemical equation. This is 1 to 2. One mole of nitrogen yielded two moles of ammonia.
At STP, one mole of a gas occupies a volume of 22.4, hence we need to know the volume occupied by a volume of 44.8L of ammonia. This is equal to 44.8/2 = 2 moles
Now we have seen the actual number of moles of ammonia yielded. Since this is the same as the theoretical, it means that only one mole of nitrogen was also used up.
Since it is one mole, the volume at STP is thus 22.4L
D. The products will have more energy than the reactants.
Answer:
Final Volume = 0.31L
Explanation:
Initial volume V1 = 1L
Initial pressure P1 = 150kPa
Initial temperature, T1 = 25'C + 273 = 298K
Final Pressure P2 = 600kPa
Final Temperature T2 = 100'C + 273 = 373K
Final Volume V2 = ?
The combined gas law is given as;
P1V1 / T1 = V2P2 / T2
150 * 1 / 298 = V2 * 600 / 373
V2 = 55950 / 178800 = 0.31L
