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3 years ago

Calculate the mass of water produced when 25.0g of oxygen gas react with

Chemistry

1 answer:

Arlecino [84]3 years ago

4 0

Reaction.-

2H2 + O2 ---> 2H2O

Given,

Mass of Oxygen - 25.0 g

Mass of Hydrogen - 42.0 g

now we have to calculate moles

moles of O2 = 25/16 = 1.5625 moles

Moles of H2 = 42/2 = 21 moles

Hydrogen gas is in excess,

now using reaction

1 mole of oxygen gas forms 2 moles of H2O

1.5625 mole of Oxygen gas will form x mole of H20

x = 3.125 moles

molar mass of H2O is 18g

mass of 3.125moles of water = 3.125 × 18 = 56.25 g.

Hence the mass of water produced when 25.0g of oxygen gas react with 42.0g of hydrogen gas is 56.25 grams

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2 years ago

2.25 g/ml at 700mmHg, 200 C. Find molar mass of gas.

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Answer:

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Given data:

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3 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

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$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

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So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

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$Total~volume=50mL+10mL=60mL$