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RideAnS [48]
3 years ago
8

If you have 6 moles of reactant A and excess of B and C, how much product E would be formed

Chemistry
1 answer:
pychu [463]3 years ago
3 0
B and C are in excess so amount of E will be determined by A. 

Amount of product is determined by limiting reagents - Always.

Hence 6 moles of E will be formed.

Hope this helps!
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How many sulfur atoms are there in 21.0 g of al2s3?
ValentinkaMS [17]

Given the mass of aluminum sulfideAl_{2}S_{3} = 21.0 g

Molar mass of Al_{2}S_{3} = 2 * Molar mass of Al + 3 *molar mass of S = 2(27g/mol)+3(32.g/mol)=150.g/mol

Calculating the moles of Al_{2}S_{3}:

21.0 g *\frac{1 mol}{150 g} = 0.14 mol Al_{2}S_{3}

Each mole Al_{2}S_{3} has 3 mol S.

Each mole S constitutes 6.022*10^{23} atoms of S

Calculating atoms of S in 0.14 mol Al_{2}S_{3}:

0.14 mol Al_{2}S_{3} *\frac{3mol S}{1 mol Al_{2}S_{3}  }*\frac{6.022*10^{23}atoms S }{1 mol S}

= 2.53 * 10^{23} atoms of S

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3 years ago
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Answer:

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Explanation:

6 0
3 years ago
10. Aqua means "water" or "a light blue color." Marine refers to the sea. Use this information
Zielflug [23.3K]

Answer:

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3 0
2 years ago
Read 2 more answers
The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces
Oxana [17]

Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>

NaN₃(s) → Na(s) + 1.5 N₂(g)

<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>

First, we have to calculate the moles of N₂ from the ideal gas equation.

P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol

The moles of NaN₃ are:

1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

1.22mol.\frac{65.01g}{mol} =79.3g

5 0
3 years ago
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Savatey [412]
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