How many grams of which reactant will remain unreacted when 10.0 g of N2H4(l) reacts with 15.0 g of N2O4(l) in the following reaction?
2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g)
Assume 100% yield.
| 4.78 g of N2H4 |
| 2.39 g of N2H4 |
| 0.01 g of N2O4 |
| 0.64 g of N2O4 |
| 0.22 g of N2H4 |
Step by Step, Thanks!
This can be done with the help
of limiting reagent and excess reagent concept.
How many grams of which reactant will remain unreacted when 10.0 g of N2H4(l) reacts with...
50.0 g N2O4 reacts with 45.0 g N2H4. Determine the percent yield if 12.5 g of N2 are actually formed. N2O4(l) + 2N2H4(l) ---> 3N2(g) + 4H2O(g)
Determine the amount of excess reactant that remains after 50.0 g N2O4 and 45.0 g N2H4 react. N2O4(l) + 2N2H4(l) ---> 3N2(g) + 4H2O(g)
Consider the reaction between N2H4 and N2O4: 2N2H4(g)+N2O4(g)?3N2(g)+4H2O(g) A reaction vessel initially contains 21.0 g N2H4 and 74.9 g of N2O4. Hint: The limiting reactant is completely consumed, but the reactant in excess is not. Use the amount of limiting reactant to determine the amount of products that form and the amount of the reactant in excess that remains after complete reaction. Part B: Calculate the mass of N2O4 that will be in the reaction vessel once the reactants have...
For the chemical reaction shown, 2H2O2(l)+N2H4(l)⟶4H2O(g)+N2(g) determine how many grams of N2 are produced from the reaction of 9.55g of H2O2 and 5.81 g of N2H4.
When 10.0 g of hydrogen gas reacts with 40.0 g of nitrogen gas, how much ammonia gas forms? Which substance is the limiting reactant? If a chemist runs the reaction knowing they should obtain a percent yield of 78.5% and must produce a total of 30.5 grams of ammonia, how many grams of nitrogen should they start with if they have an excess amount of hydrogen?
For the reaction shown, calculate how many moles of NH3NH3 form when each amount of reactant completely reacts. 3N2H4(l)→4NH3(g)+N2(g) 1. a. 4.9 mol N2H4 b. 2.23 mol N2H4 c. 6.3×10^−2 mol N2H4 d. 9.48×107 mol N2H4
How many grams of lead (II) iodide would be produced based on
the amount of lead (II) nitrate used?
What is the theoretical yield of lead (II) iodide (in grams) for
this reaction based on the limiting reactant?
What was the actual yield (in grams) of the solid for this
experiment?
What was the percent yield of the reaction?
How many moles of iodide were left unreacted?
How many moles of NaI were left unreacted?
How many grams...
You may want to reference (Pages 141-145) Section 4.2 while completing this problem. Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation 3 N2H4(l) → 4 NH3(g)+N2(g) Part A 2.5 moles N2H4Part B 3.55 mol N2H4 Part C 61.3 g N2H4 Part D 4.58 kg N2H4
Which substance is the limiting reactant when 16.0 g of sulfur reacts with 10.0 g of oxygen and 12.0 g of sodium hydroxide according to the following chemical equation? 2 S(s) + 3 O2(g) + 4 NaOH(aq) → 2 Na2SO4(aq) + 2 H2O(l)
How many grams of NH3 form when 24.3 L of H2(g) (measured at STP) reacts with N2 to form NH3 according to this reaction? N2(g)+3H2(g)→2NH3(g)