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Elena L [17]
3 years ago
6

Ammonia is produced by the following reaction. 3H2(g) + N2(g) ---> 2NH3(g) When 7.00 g of hydrogen react with 70.0 g of nitro

gen, hydrogen is considered the limiting reactant because
Chemistry
2 answers:
ycow [4]3 years ago
6 0
To determine the limiting reactant, we first calculate the number of moles of the reactant.

7.00 g H2 ( 1 mol / 2.02 g ) = 3.47 mol H2
70.00 g N2 ( 1 mol / 28.02 g) = 2.50 mol N2

From the balanced reaction, we see that there is a 3 is to 1 ratio of the reactants. So, for every 3 moles of H2 we need 1 mole of N2. Given the amount of reactant, for 2.50 moles of N2 we need 7.5 moles of H2 which obviously we have a lesser amount. Hydrogen is the limiting reactant.
Sveta_85 [38]3 years ago
6 0
The correct answer is A
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Equation of reaction
Hcl+NaoH-->Nacl+H2O
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Using the formula CaVa/CbVb=na/nb
Ca(Concentation of acid)= 0.100M
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VB=50.00ml
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from the formula
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Va= 0.2*50*1/0.1*1
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2 years ago
I need the answer today!
olchik [2.2K]

_____________________________________________

<h2 /><h2 /><h2><u>Solution </u><u>3</u><u> Is The Most Concentrated</u></h2>

<h3>S1:</h3>

M = m/v

= 100ml ÷ 2 spoons × 100%

= - 5,000 μg/ppb³

= <u>50% Diluted</u>

<h3>S2:</h3>

M = m/v

= 200ml ÷ 5 spoons × 100%

= - 4,000 μg/ppb³

= <u>40% Diluted</u>

<h3>S3:</h3>

M = m/v

= 300ml ÷ 6 spoons × 100%

= - 5,000 μg/ppb³

= <u>50% Diluted</u>

<h3>S4: </h3>

M = m/v

= 600ml ÷ 8 spoons × 100%

= - 75,000 μg/ppb³

= <u>75% Diluted</u>

_____________________________________________

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2 years ago
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