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Vaselesa [24]
3 years ago
11

A balanced equation tells you the mole ratio of all reactants and products. Using the chemical reaction below how many moles of

NH3 are made when 3 moles of hydrogen gas are used?
2N2 + 3H2 ---> 2NH3
Chemistry
2 answers:
Hunter-Best [27]3 years ago
6 0
From  the  reaction  above    2  moles  of  N2  reacted  with  3  moles  of  H2  to  give  2  moles of   ammonia
THese  implies  that   the   mole  ratio  of  hydrogen  to  ammonia   is   3:2
Therefore  if   3  moles  of  hydrogen   were  used  the  moles  of  ammonia  is  therefore
(3 x2)/3=  2moles
vichka [17]3 years ago
5 0
<span>By using the mole ratio, we can determine that 2 moles of NH3 are made when 3 moles of hydrogen gas are present. The numbers in front of the chemicals tell us the relative amounts consumed and produced. Since there is a 3 in front of H2 and a 2 in front of NH3, this tells us that for every 3 moles of H2 gas used, 2 moles of NH3 are made.</span>
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A.) A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. If the titration required 1
nikklg [1K]

Answer:

A) 0.1225 M

B) 100.4 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

17.73 mL of 0.1036 M NaOH react. The reacting moles are:

0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.

Step 4: Calculate the molar concentration of HA

1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:

M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M

Step 5: Calculate the molar mass of HA

1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:

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5 0
3 years ago
Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord
Marina CMI [18]

Answer:

M_{acid}=0.563M

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the acid to obtain:

M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M

Regards!

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Answer is: D- there are different elements as products.
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