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MrRa [10]
3 years ago
13

How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?

Chemistry
1 answer:
vova2212 [387]3 years ago
7 0

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

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What mass of sulfur has to burn to produce 4.5L SO2 at 300°C and 101 kPa in the following reaction?s(s)+O2(g) SO,(g)
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<span>Answer: option B. 3.07 g

Explanation:

1) given reaction:

S(s) + O₂ (g) → SO(g)

2) Balanced chemical equation:

</span><span>2S(s) + O₂ (g) → 2SO(g)

3) Theoretical mole ratios:

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3) number of moles of 4.5 liter SO₂ at</span><span> 300°C and 101 kPa

use the ideal gas equation:

pV = nRT

with V = 4.5 liter
p = 101 kPa
T = 300 + 273.15 K = 573.15 K
R = 8.314 liter×kPa / (mol×K)

=> n = pV / (RT) =

n =  [101 kPa × 4.5 liter] / [8.314 (liter×kPa) / (mol×K)  × 573.15 K ]

n = 0.0954 mol SO

4) proportion with the theoretical ratio S / SO

 2 mol S                   x
-------------- = ----------------------
 2 mol SO      0.0954 mol SO

=> x = 0.0954 mol S.

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mass = number of moles × atomic mass

mass = 0.0954 mol × 32.065 g/mol = 3.059 g of S

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</span>
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