Your answer is False I think
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
Answer:
Explanation:
First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)
Then we solve the ecuation for ΔH°reaction
ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)
ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol
ΔH°reaction = -1.41 *10^3 kJ/mol
B, millimeters because paper is really thin therefore it'd require small measurements for units.
