Answer:
204g of NH3
Explanation:
The balanced equation for the reaction is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the number of mole NH3 produced by reacting 6moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 6 moles of N2 will react to produce = 6 x 2 = 12 moles of NH3.
Finally, we shall convert 12 moles of NH3 to grams. This is illustrated below:
Number of mole of NH3 = 12 moles.
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 =..?
Mass = mole x molar mass
Mass of NH3 = 12 x 17
Mass of NH3 = 204g.
Therefore, 204g of NH3 will be produced from the reaction.
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
Answer:
Transition Metals
Explanation:
The elements in groups 3-12 are called Transition Metals. These groups contain metals that usually form multiple cations. All other groups on the table (1, 2, 13-18) are called Main Group Elements.
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
