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Bond [772]
3 years ago
11

WILL MARK BRAINLIEST

Chemistry
1 answer:
tiny-mole [99]3 years ago
5 0

Answer:

b) 590

Explanation:

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A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat
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Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

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571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

Mass percentage of H2SO4 (%w/w) is 42.99 %

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1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)
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Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

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