Answer:
Q = mcT ...you can either substitute the molar heat capacity of water in the place of c or the specific heat capacity of water.
Explanation:
Answer:
0.41kg/sec
Explanation:
PV= nRT
Given : V= 505 L
P=0.88 atm
R= 0.08206 Latm/K*mol
T= 172 .0C = 172+273 = 445 K
n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed
As per reaction : N2 + 3H2 ----> 2NH3
1 mole N2 is consumed to produce 2 moles NH3
moles of NH3 produced per sec :
(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec
grams of NH3 produced per sec =
24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec
rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec
= 0.41kg/sec
Answer:
40.5 g of P₄O₁₀ are produced
Explanation:
We state the reaction:
P₄ + 5O₂ → P₄O₁₀
We do not have data from P₄ so we assume, it's the excess reactant.
We need to determine mass of oxygen and we only have volumne so we need to apply density.
Density = mass / volume, so Mass = density . volume
Denstiy of oxygen at STP is: 1.429 g/L
1.429 g/L . 16.2L = 23.15 g
We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles
5 moles of O₂ can produce 1 mol of P₄O₁₀
Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles
We determine the mass of product:
0.138 mol . 292.88 g/mol = 40.5 g
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