C . oxygen
i had this the other day
Answer:
chuma ya ndoshi.............
Explanation:
.
kwenda ukalale .............
Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
![x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:](https://tex.z-dn.net/?f=x%20%5C%3A%20mol%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%5Cfrac%7B2%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%2035%7D%7B1000%7D%20%20%5C%5C%20%20%3D%200.07%20%5C%3A%20moles%20%5C%3A%20)
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
![= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles](https://tex.z-dn.net/?f=%20%3D%20%20%5Cfrac%7B2%7D%7B6%7D%20%20%5Ctimes%200.07%20%5C%5C%20%20%5C%3A%20%20%3D%200.0233333%20%5C%3A%20moles)
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
![grams \: = moles \: \times \: rfm](https://tex.z-dn.net/?f=grams%20%5C%3A%20%20%3D%20moles%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%20rfm)
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
![= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams](https://tex.z-dn.net/?f=%20%3D%200.0233333%20%5C%3A%20moles%20%20%5C%3A%20%20%5Ctimes%20%20%5C%3A%2026.982%20%5C%3A%20%20%5Cfrac%7Bg%7D%7Bmol%7D%20%20%5C%5C%20%20%3D%200.625799%20%5C%3A%20grams)
The number of grams of aluminum required to react with HCl is 0.6258 g.
Answer: The amount of water produced is 9.3 grams
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
![CH_4+2O_2\rightarrow CO_2+2H_2O](https://tex.z-dn.net/?f=CH_4%2B2O_2%5Crightarrow%20CO_2%2B2H_2O)
mass of reactants = mass of methane + mass of oxygen = 22.5 g + 35.7 g = 58.2 g
mass of products = mass of carbon dioxide + mass of water = 48.9 g + mass of water
48.9 g + mass of water = 58.2 g
mass of water = 9.3 g