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
Magnesium + oxygen makes Magnesium Oxide.
The big numbers are for balancing the equation.
D. is the missing element.
Carlos <em>hasn’t done research</em> on what other scientists have observed and investigated on this topic. If he hasn’t done his literature research, he may just be repeating the experiments of other scientists working in the same area.
Options “A.”, “B.”, and “C.” are all part of the scientific method.
Answer:
5.38 moles of CO₂ are produced
Explanation:
This is the reaction:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
First of all, let's convert the mass of C₃H₈ to moles (mass / molar mass)
79 g / 44 g/mol = 1.79 moles
So ratio is 1:3.
1 mol of C₃H₈ is needed to produce 3 moles of CO₂
1.79 moles of C₃H₈ would produce (1.79 .3) /1 = 5.38 moles