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romanna [79]
3 years ago
15

If 3.0 moles of x and 4.0 moles of y react according to the hypothetical reaction below, how many moles of the excess reactant w

ill be left over at the end of the reaction? x + 2 y yields xy2
Chemistry
1 answer:
Lesechka [4]3 years ago
6 0
Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.

the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x 
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.


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What is the ph of an aqueous solution with the hydronium ion concentration : [h3o+] = 3 x 10-5 m ?
BaLLatris [955]

The concentration of hydrogen can be shown as:

[H+ ] = 3 * 10-5 M

pH can be determined as:

pH = - log [H+ ]

= - log (3 * 10-5)

= 4.53

Thus the pH of solution is 4.53


8 0
3 years ago
I need the answer for this question
ch4aika [34]
I want to say A only... Hope I helped!!
5 0
3 years ago
Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0
3 years ago
D5W is a solution used as an intervenous fluid.It is 5.0% by mass solution of the dextrose (C6H12O6) in water. if the density of
Alona [7]

Answer:

0.29 mol/L

Explanation:

Its density is 1.029 g/ml so in a liter (1000 mL) there is 1029 g of solution, but only 5% is dextrose.

0.05x1029=51.45

So in a liter of D5W solution there is 51.45 g of dextrose.

Dextrose molar mass iss 180.156 g/mol, so in 51.45 g of dextrose there is

51.45/180.156=0.29 mol

In one liter of solution there is 0.29 mol of dextrose, so the molarity of such solution is 0.29 mol/L.

6 0
3 years ago
Very Important!! Electron lender or borrower?
worty [1.4K]
Sodium lends 1 electron.
Phosphorus borrows 3 electrons.
Potassium lends one electron.
Oxygen borrows 2 electrons.
Iodine borrows one electron.
Cesium lends 1 electron.
Bromine borrows 1 electron.
Sulfur borrows 2 electrons.
And magnesium lends 2 electrons.
4 0
4 years ago
Read 2 more answers
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