The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
PV = nRT
If pressure increases, so will moles.
The answer is 2) Increase.
Answer:
171.34 g/mol
Explanation:
Ba molar mass = 137.328 g/mol
O molar mass = 15.999 g/mol * 2 = 31.9980 g/mol
H molar mass = 1.008 g/mol * 2 = 2.0160 g/mol
137.328 + 31.9980 + 2.0160 = 171.3420 = 171.34 g/mol
It was empty , dark , and cold
Answer:
83.64%.
Explanation:
∵ The percent yield = (actual yield/theoretical yield)*100.
actual yield of CO₂ = 2300 g.
- We need to find the theoretical yield of CO₂:
For the reaction:
<em>CH₄ + 2O₂ → 2H₂O + CO₂,</em>
1.0 mol of CH₄ react with 2 mol of O₂ to produce 2 mol of H₂O and 1.0 mol of CO₂.
- Firstly, we need to calculate the no. of moles of 1000 g of CH₄ using the relation:
<em>no. of moles of CH₄ = mass/molar mass</em> = (1000 g)/(16.0 g/mol) = <em>62.5 mol.</em>
<u><em>Using cross-multiplication:</em></u>
1.0 mol of CH₄ produces → 1.0 mol of CO₂, from stichiometry.
∴ 62.5 mol of CH₄ produces → 62.5 mol of CO₂.
- We can calculate the theoretical yield of carbon dioxide gas using the relation:
∴ The theoretical yield of CO₂ gas = n*molar mass = (62.5 mol)(44.0 g/mol) = 2750 g.
<em>∵ The percent yield = (actual yield/theoretical yield)*100.</em>
actual yield = 2300 g, theoretical yield = 2750 g.
<em>∴ the percent yield</em> = (2300 g/2750 g)*100 = <em>83.64%.</em>
