18. A 35.0 G piece of metal wire is heated and the temperature changes from 21°C to 52°C the specific heat of the metal is 0.900
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<u>Answer:</u> The amount of formed is 469.8 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
......(1)
Given mass of phosphine = 225 g
Molar mass of phosphine = 34 g/mol
Putting values in equation 1, we get:
The given chemical reaction follows:
Assuming that oxygen gas is present in excess, it is considered as an excess reagent.
Phosphine is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
4 moles of phosphine produces 1 mole of
So, 6.62 moles of phosphine will produce = of
Now, calculating the mass of by using equation 1:
Molar mass of = 283.9 g/mol
Moles of = 1.655 moles
Putting values in equation 1, we get:
Hence, the amount of formed is 469.8 grams.
The reaction is as follows:
2 H₂(g) + O₂(g) → 2 H₂O(g) . ΔH = - 483.5 kJ
Using the change in enthalpy and heat, calculate the moles as follows:
Moles of H₂ = -
x 2 mol H₂
= - 216 kJ / (-483.5 kJ) x 2 mol H₂
= 0.893 mol H₂
Calculate the mass of H₂ using the moles and molar mass as follows:
0.893 mol H₂ x (2.02 g H₂ / 1 mol H₂) = 1.79 g H₂
Therefore, the mass of hydrogen gas is 1.79 g
2 H₂(g) + O₂(g) → 2 H₂O(g) . ΔH = - 483.5 kJ
Using the change in enthalpy and heat, calculate the moles as follows:
Moles of H₂ = -
= - 216 kJ / (-483.5 kJ) x 2 mol H₂
= 0.893 mol H₂
Calculate the mass of H₂ using the moles and molar mass as follows:
0.893 mol H₂ x (2.02 g H₂ / 1 mol H₂) = 1.79 g H₂
Therefore, the mass of hydrogen gas is 1.79 g