Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
Answer:
C
Explanation:
It is letter C since you are currently experimenting to see if by the amount of rainfall it gets it can <em><u>grow</u></em><em><u> </u></em><em><u>more</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>less</u></em>. So you control the amount of rain
Answer:
8 grams
Explanation:
The molar mass of CH4 is about 16 grams per mol. So you get the answer by mutlitplying 0.5 by 16, (cancel out moles by placing 1 on the bottom of the fraction).
The symbols indicate the physical state of each reactant
Grass is biotic- living.
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