Aluminum hydroxide 
can behave as a base and neutralize sulfuric acid 
as in the following equation:
(Balanced)
(a)
. Thus the ratio between the number of moles of the two reactants available:
![n(\text{Al}(\text{OH})_3, \text{supplied}) / n(\text{H}_2\text{SO}_4, \text{supplied})\\= [m(\text{Al}(\text{OH})_3)/ M(\text{Al}(\text{OH})_3)] / [n(\text{H}_2\text{SO}_4) / M(\text{H}_2\text{SO}_4)]\\= [23.7 / (26.98 + 3 \times(16.00 + 1.008))]/[29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00)]\\\approx 1.01](https://tex.z-dn.net/?f=n%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%2C%20%5Ctext%7Bsupplied%7D%29%20%2F%20n%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%2C%20%5Ctext%7Bsupplied%7D%29%5C%5C%3D%20%5Bm%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%29%2F%20M%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%29%5D%20%2F%20%5Bn%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%29%20%2F%20M%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%29%5D%5C%5C%3D%20%5B23.7%20%2F%20%2826.98%20%2B%203%20%5Ctimes%2816.00%20%2B%201.008%29%29%5D%2F%5B29.5%20%2F%20%282%20%5Ctimes%201.008%20%2B%2032.07%20%2B%204%20%5Ctimes%2016.00%29%5D%5C%5C%5Capprox%201.01)
The value of this ratio required to lead to a complete reaction is derived from coefficients found in the balanced equation:
The ratio for the complete reaction is smaller than that of the reactants available, indicating that the species represented on the numerator, , is in excess while the one on the denominator, , serves as the limiting reagent.
(b)
The quantity of water produced is dependent on the amount of limiting reactants available. 
of sulfuric acid is supplied in this reaction as the limiting reagent. 
moles of water molecules are produced for every 
moles of sulfuric acid consumed. The reaction would thus give rise to 
of water molecules, which have a mass of 
.
(c)
(d)
The quantity of , the reactant in excess, is dependent on the number of moles of this species consumed in the reaction and thus the quantity of the limiting reagent available. The consumption of every moles of sulfuric acid, the limiting reagent, removes 
moles of aluminum hydroxide from the solution. 
of sulfuric acid is initially available as previously stated such that 
, or 
, of would be eventually consumed.
of would thus be in excess by the end of the reaction process.
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Answer:
35 minutes
Explanation:
The flow rate of the stream is the rate of volume of water with respect to time. It expresses the volume of water flowing per unit time.
Flow rate = 
From the given question,
average flow rate of the stream = 420 cubic meters per minute, and the required volume = 14700 cubic meters.
So that;
time = 
= 
= 35
time = 35 minutes
Therefore, with the given average flow rate, the steam will produce 14,700 cubic meters of water in 35 minutes.
The mass of CH₄ that is consumed is 10.598 g.
<h3>What is enthalpy?</h3>
The enthalpy of a system is defined as the sum of the internal energy of a system and the energy that is produced due to its pressure and volume.
It is given by
H = U+PV
The following reaction takes place during the combustion of CH₄
Energy evolved during the combustion, E = 587.9 kJ
Enthalpy of combustion of methane, ∆H = 890.3 kJ
= 587.9 /890
Mass of CH₄ that is consumed = no of moles × molar mass of CH₄
= 587.9 /890 × 16
= 10.598 g
The mass of CH₄ which is consumed is 10.598 g
Learn more about enthalpy:
brainly.com/question/14047927
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When aluminum metal is made to contact with chlorine gas (Cl₂), a highly exothermic reaction proceeds. This produces aluminum chloride (AlCl₃) powder. The balanced chemical equation for this reaction is shown below:
2Al(s) + 3Cl₂(g) → 2AlCl₃(s)
Since it was stated that aluminum is in excess, this means that the amount of AlCl₃ produced will only depend on the amount of Cl₂ gas available. The molar mass of Cl₂ is 70.906 g/mol. Using stoichiometry, we have the following equation:
(21.0 g Cl₂/ 70.906 g/mol Cl₂) x 2 mol AlCl₃/ 2 mol Cl₂ = 0.1974 mol AlCl₃
Thus, we have determined that 0.1974 <span>moles of aluminum chloride can be produced from 21.0 g of chlorine gas. </span>
