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vitfil [10]
3 years ago
12

How many moles of HCl would react with 37.1 mL of 0.138 M Sr(OH)2

Chemistry
1 answer:
Lena [83]3 years ago
5 0

Answer is: 0.102 moles of HCl would react.

Balanced chemical reaction:

2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).

V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.

V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.

c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.

n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).

n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.

n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.

From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.

n(HCl) = 2 · n(Sr(OH)₂).

n(HCl) = 2 · 0.0051 mol.

n(HCl) = 0.0102 mol; amount of the hydrochloric acid.

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alexandr402 [8]

Answer:

x_{N_2F_2}= 0.415\\\\x_{SF_6}=0.585

Explanation:

Hello!

In this case, since the mole fraction of both gases in the tank is computed via:

x_{N_2F_2}=\frac{n_{N_2F_2}}{n_{N_2F_2}+n_{SF_6}} \\\\x_{SF_6}=\frac{n_{SF_6}}{n_{N_2F_2}+n_{SF_6}}

It means we need to compute the moles of each gas, just as it is shown down below:

n_{N_2F_2}}=5.53gN_2F_2*\frac{1molN_2F_2}{66.01gN_2F_2} =0.0838molN_2F_2\\\\n_{SF_6}=17.3gSF_6*\frac{1molSF_6}{146.06gSF_6} =0.118molSF_6

Thus, the mole fractions turn out:

x_{N_2F_2}=\frac{0.0838mol}{0.0838mol+0.118mol}= 0.415\\\\x_{SF_6}=\frac{0.0838mol}{0.0838mol+0.0838mol}=0.585

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