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3 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP

Chemistry

2 answers:

Dvinal [7]3 years ago

7 0

Vol.250 before its to much pressure

alisha [4.7K]3 years ago

5 0

Answer: 3808 ml

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.03}{24}=0.17moles$

$2Mg+O_2\rightarrow 2MgO$

2 moles of magnesium react with= [tex[2\times 22.4=44.8L[/tex] of oxygen at STP

Thus 0.17 moles of magnesium react with= $\frac{44.8}{2}\times 0.17=3.8L=3808ml$

Thus the volume of oxygen required to react with 4.03 g of magnesium at STP is 3808 ml.

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A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titra

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Answer:

See explanation below

Explanation:

In order to calculate this, we need to use the following expression to get the concentration of the base:

MaVa = MbVb (1)

We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:

Atomic weights of the elements to be used:

K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol

MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol

Now, let's calculate the mole of KHP:

moles = 0.5053 / 204.2189 = 0.00247 moles

With the moles, we also know that:

n = M*V (2)

Replacing in (1):

n = MbVb

Now, solving for Mb:

Mb = n/Vb (3)

Finally, replacing the data:

Mb = 0.00247 / (13.4473/1000)

Mb = 0.184 M

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What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?

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Answer:

$M=0.638M$

Explanation:

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In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

$n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol$