M Fe = 55,85g ≈ 56g
m O = 16g
m N = 14g
m Fe(NO3)3 = 56g + 3*14g + 3*3*16g = 242g/mol
answer: (4)
Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
<span>You need to have NAD+ as a source of oxidation for the pyruvate, as well as a supply of coenzyme A. CO2 is released by the pyruvate as a carboxyl group is removed</span>
Answer:
The limiting reactant is H₂
Explanation:
The reaction of hydrogen (H₂) and carbon monoxide (CO) to produce methanol (CH₃OH) is the following:
2H₂(g) + CO(g) → CH₃OH(g)
From the balanced chemical equation, we can see that 1 mol of CO reacts wIth 2 moles of H₂. So, the stoichiometric ratio is:
2 mol H₂/1 mol CO = 2.0
We have 500 mol of CO and 750 mol of H₂, so we calculate the ratio to establish a comparison:
750 mol H₂/500 mol CO = 1.5
Since 2.0 > 1.5, we have fewer moles of H₂ than are needed to completely react with 500 moles of CO. In fact, we need 1000 moles of H₂ and we have 750 moles. So, the limiting reactant is H₂.
