Answer:
0.52 mol
Explanation:
Using the general gas equation formula:
PV = nRT
Where;
P = pressure (atm)
V = volume (Liters)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At STP (standard temperature and pressure), temperature of a gas is 273K, while its pressure is 1 atm
Using PV = nRT
n = PV/RT
n = (1 × 11.74) ÷ (0.0821 × 273)
n = 11.74 ÷ 22.41
n = 0.52 mol
There are 0.52 moles in the basketball
Answer:
101.50 g H₂O
Explanation:
The mole ratio of HNO₃ and H₂O is 6 : 2
Hence, 16.9 moles of HNO₃ will produce = 2/6×16.9 = 5.63 moles of H₂O
Also,
Mass = Moles × M.Mass
Mass = 5.63 mol × 18.02 g/mol
Mass = 101.50 g H₂O
The correct answer is 2.53 g of precipitate, BaCrO4.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

Regards.
Answer:
CuSO4
Explanation:
Na2S + CuSO4 → Na2SO4 + CuS
The reaction is balanced (same number of elements in each side)
To determine limiting reagent you need to know the moles you have of each.
Molar mass Na2S = 23 * 2 + 32 = 78
Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5
Na2S mole = 15.5 / 78 = 0.2
CuSO4 mole = 12.1/159.5 = 0.076
*Remember mole = mass / MM
With that information now you have to divide each moles by its respective stoichiometric coefficient
Na2S stoichiometric coefficient : 1
Na2S : 0.2 / 1 = 0.2
CuSO4 stoichiometric coefficient: 1
CuSO4: 0.076 / 1 = 0.076
The smaller number between them its the limiting reagent, CuSO4