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3 years ago

You have 16.7 grams of hydrogen and 15.4 grams of oxygen in a synthesis rxn. Which is the limiting reagent?

Chemistry

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

oxygen is limiting reactant

Explanation:

Given data:

Mass of hydrogen = 16.7 g

Mass of oxygen = 15.4 g

Limiting reactant = ?

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 16.7 g/ 2 g/mol

Number of moles = 8.35 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 15.4 g/ 32 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of both reactant with product,

H₂ : H₂O

2 : 2

8.35 : 8.35

O₂ : H₂O

1 : 2

0.48 : 2×0.48 = 0.96 mol

The number of moles of water produced by oxygen are less so it will limiting reactant.

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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 l of carbon dioxide at 20.0 Â°c an

kvv77 [185]

Answer:

Percentage by mass of calcium carbonate in the sample is 93.58%.

Explanation:

Assumptions:

calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
the sample did not contain any other compound that released or reacted with carbon dioxide.

$PV = nRT$

$n = \frac{PV}{RT}$

$P = 791 mmHg = \frac{791 mmHg}{760 mmHg} x 1 atm = 1.040789 atm.$

$R = 8.314 \frac{J}{K*mol} = 0.082 \frac{L*atm}{K*mol}$

$T = 20^{0}C + 273^{0} C = 293 K$

$n = \frac{1.040789 atm * 1.14 L}{0.082 \frac{L * atm}{K * mol} * 293 K }$

$n = 0.0493839 mol.$

given the equation of the reaction:

$CaCO_{3} + 2HCL = CaCl_{2} + CO_{2} + H_{2}O$

from assumption:

amount carbon dioxide = amount of calcium carbonate

$n(CaCO_{3} ) = n(CO_{2} )$

reacting mass ( m ) = Molar Mass ( M ) * Amount ( n )

$m(CaCO_{3} ) = n(CaCO_{3} ) * M(CaCO_{3} )$

m = 0.04938397 $mol$ * 100 $\frac{g}{mol}$ = 4.93839 $g$

percentage by mass of CaC $O_{3}$ = $\frac{mass of pure }{mass of impure}*100 = \frac{4.93839g}{5.28g}*100 = 93.5802%$ .

8 0

2 years ago

Craig needs to organize, analyze, and interpret large amounts of quantitative data that he generated during an experiment. Which

notsponge [240]

Answer:

C. a computer

Explanation:

Quantitative data are measures of values or counts and are expressed as numbers. The best tool out of all shown to deal with a large amount of numerical data is a computer

4 0

2 years ago

What mass of CO2 in grams will form 8.94 g O2 completely react?

VLD [36.1K]

Through ratio and proportion, for every mole of O2, there is also a mole equivalent of CO2. In this case, we divide first 8.94 g O2 by 32 g/mol to convert the mass to mole. That is equivalent to 0.279375 mol which is also the amount of CO2. Multiplied by 44g/mol which is the molar mass of CO2, the answer is 12.29 grams CO2.

8 0

3 years ago

What occurs along a convergent plate boundary

HACTEHA [7]

At convergent plate boundaries, oceanic crust is often forced down into the mantle where it begins to melt. Magma rises into and through the other plate, solidifying into granite, the rock that makes up the continents. Thus, at convergent boundaries, continental crust is created and oceanic crust is destroyed.

8 0

3 years ago

Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s

slava [35]

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

3 0

3 years ago