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.
2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008
When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4%
For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal
When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:
(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
Answer:
The molar mass of Mg(NO₃)₂, 148.3 g/mol.
Explanation:
Step 1: Given data
- Mass of Mg(NO₃)₂ (solute): 42.0 g
- Volume of solution: 259 mL = 0.259 L
Step 2: Calculate the moles of solute
To calculate the moles of solute, we need to know the molar mass of Mg(NO₃)₂, 148.3 g/mol.
42.0 g × 1 mol/148.3 g = 0.283 mol
Step 3: Calculate the molarity of the solution
M = moles of solute / liters of solution
M = 0.283 mol / 0.259 L
M = 1.09 M
A balanced equation must have the same number of atoms on the both sides of equation.
