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viva [34]
3 years ago
11

Please help me this is my fourth attempt.

Chemistry
1 answer:
Vaselesa [24]3 years ago
8 0

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

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C. they both eat and grow a lot, i hope it helps

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Balance the equation <br><br> __C7H6O2 + __O2 —&gt; __CO2 + __H2O
ser-zykov [4K]

2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

<u>Explanation:</u>

C₇H₆O₂ + O₂ → CO₂ + H₂O

First we have to balance the O- atoms, we have to put 6 in front of water so there are 12 H atoms on RHS, to balance it we need to put 2 in front of  C₇H₆O₂, and so we have 14 C - atoms on LHS, 28 + 6 = 34 O - atoms on RHS, so we have to put 15 in front of Oxygen in LHS, so that each and every atom in the equation gets balanced now. The balanced equation is,  2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

4 0
3 years ago
Dinitrogen oxide (N₂O) gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet ga
RideAnS [48]

Answer:

126.73 mL

Explanation:

The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:

750 = 18.85 + p

p = 750 - 18.85

p = 731.15 torr

It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that

\frac{p1V1}{T1} = \frac{p2V2}{T2}

Where <em>p</em> is the pressure, <em>V</em> is the volume, <em>T</em> is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:

p1V1 = p2V2, so:

731.15x130 = 750xV2

750V2 = 95049.5

V2 = 126.73 mL

4 0
3 years ago
The graph indicates the running route for Tobias.
Musya8 [376]

Answer: b

Explanation:

7 0
3 years ago
Read 2 more answers
6 NaOH + 2 Al ???? 2 Na3AlO3 + 3 H2 How much aluminum is required to produce 17.5 grams of hydrogen? How many moles of NaOH are
ANTONII [103]

Answer:

a) 157.5 grams of aluminum.

b) 1 mol

c) 9 g

Explanation:

The reaction is :

6 NaOH + 2Al ---> 2 Na_{3}AlO_{3} + 3H_{2}

As per balanced equation

a) 3 moles of hydrogen will be produced from two moles of aluminium.

The atomic mass of aluminium = 27

therefore

3X2 grams of hydrogen is produced from 2 X 27 grams of Al

1 gram of hydrogen will be produced from \frac{2X27}{3X2}= 9g

therefore 17.5 will be produced from = 9X 17.5 = 157.5 grams of aluminum.

b) as per balanced equation three moles or six gram of hydrogen is produced from 6 moles of NaOH.

Therefore 1 g of hydrogen will be produced from =\frac{6}{6}

or 1 gram will be prepared from = 1 mole

c) from balanced equation three moles are produced from two moles of Al (27X2 = 54 g).

thus from 54  grams gives 6 grams of hydrogen

1 grams will give = \frac{54X1}{6}= 9 g

7 0
3 years ago
Read 2 more answers
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