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2 years ago

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1) How many moles of iron can be made from 3 moles of Fe2O3?

Chemistry

1 answer:

enot [183]2 years ago

8 0

ANSWER - 6 MOLES OF IRON

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1)

One moles of Fe2O3 forming 2 moles of Fe

3 moles of Fe2O3 will form 2×3 = 6 moles of iron

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Which one is an expression of distance in Sl units?

Svetllana [295]

Answer:

500 meters

Explanation:

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2 years ago

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

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.

Moles of HCl

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

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

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;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0

2 years ago

How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC

sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

$3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2$

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

$PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L$