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Likurg_2 [28]
4 years ago
15

Consider the reaction between HCl and O2: 4HCl grams + O2 - 2 H2O liquid + 2 Cl2 grams When 63.1 grams of HCl react with 17.2 g

of O2, 49.3 grams of Cl2 are collected. What is the limiting reactant?
Chemistry
1 answer:
raketka [301]4 years ago
6 0
The balanced equation for the above reaction is as follows;
4HCl + O₂ ---> 2H₂O + 2Cl₂
stoichiometry of HCl to O₂ is 4:1
number of HCl moles present - 63.1 g / 36.5 g/mol = 1.73 mol
number of O₂ moles present - 17.2 g / 32 g/mol = 0.54 mol
if HCl is the limiting reactant; then number of O₂ moles = 1.73/4 = 0.43 mol
There are 0.54 mol present and only 0.43 mol are required to react. Therefore HCl is the limiting reactant and O₂ is in excess.
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3 years ago
ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with
Kaylis [27]

Answer:

0.57 moles (NH4)3PO4 (2 sig. figs.)

Explanation:

To quote, J.R.

"Note:  liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.

H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O  

Assuming that H3PO4 is not limiting, i.e. it is present in excess

1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"

3 0
3 years ago
According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10
wel

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg

8 0
3 years ago
What is the empirical formula of a compound with a % composition of 40.1% sulfur and 59.9% oxygen?
olga_2 [115]

Answer:

The empirical formula is SO

Explanation:

The empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

Given 40.1% Sulphur and 59.9% oxygen.

We have to assume the mass of the compound to make our calculations easy.

Let's assume that the mass of the compound is 100g, that the mass of sulphur will be 40.1g and the mass of oxygen will then be 59.9g.

Number of moles= reacting mass/ molar mass

Molar mass of Sulphur = 32g/mol

Molar mass of Oxygen = 16g/mol

No of moles of S= 40.1g/100g

=0.401 moles of S

No of moles of O = 59.9g/100g

=0.599 moles of O

•These are the relative mole ratios for the compound,

•They need to be converted from decimals into whole numbers

•Turn mole ratio into whole number ratio by dividing by all the elements by the least/smallest number of moles calculated.

Number of moles of S = 0.401moles/0.401 = 1 mol S

Number of moles of O = 0.599moles/0.401 = 1.49 mol O which is approximately 1 (to the nearest whole number, considering it tenths' value which is 4 and less than 5)

The empirical formula is therefore SO.

4 0
4 years ago
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