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VMariaS [17]
3 years ago
12

2 H2 + O2 → 2 H2O How many grams of H2O are produced when 2.50 moles of oxygen are used?

Chemistry
1 answer:
vodomira [7]3 years ago
8 0

Answer:

90g

Explanation:

The equation for the reaction is given below:

2H2 + O2 → 2H2O

From the equation above,

1 mole of O2 produced 2 moles of H2O.

Therefore, 2.5 moles of O2 will produce = 2.5 x 2 = 5 moles of H2O.

Now let us convert 5 moles to H2O to grams to obtain the desired result. This is shown below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 5 moles

Mass of H2O = ?

Mass = number of mole x molar Mass

Mass of H2O = 5 x 18 = 90g

Therefore, 90g of H2O are produced

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Explanation:

14. 13.2/10 = 1.32

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Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam
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Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of [OH^-] be x

Initial concentration of [CH3NH_2], c = 0.230 M

        CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-

at eq'm  c-x                         x                    x

Expression of K_b:

K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}

Since ,methyl-amine is a weak base,c>>x so c-x\approx c.

K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}

Solving for x, we get:

x=1.07\times 10^{-2} M

Given, HCl with 0.230 M , it dissociates fully in water which means [H^+] = 0.230 M

[OH^-]=[H^+] will result in neutral solution, since [OH^-]

Remaining [H^+] after neutralizing [OH^-]ions

[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M

pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65

The pH at the equivalence point for the titration will be 0.65.

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