Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Answer: (1)CaSO4 -> (2)O2 + (1)CaS
Explanation: edge 2020 chem
Answer:0.300M
Explanation:1) Data:
a) Initial solution
M = 1.50M
V = 50.0 ml = 0.050 l
b) Solvent added = 200 ml = 0.200 l
2) Formula:
Molarity: M = moles of solute / volume of solution is liters
3) Solution:
a) initial solution:
Clearing moles from the molarity formula: moles = M × V
moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol
b) final solution:
i) Volumen of solution = 0.050 l + 0.200l = 0.250l
ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer
Answer:
C
Explanation:
the enthalpy of reaction is independent of the reaction path
