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2 years ago

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If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?

1 answer:

vfiekz [6]2 years ago

3 0

Answer: 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles$

$3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4$

$Na_2SO_4$ is the limiting reagent as it limits the formation of product and $H_3PO_4$ is the excess reagent.

According to stoichiometry :

3 moles of $Na_2SO_4$ produce = 3 moles of $H_2SO_4$

Thus 0.0076 moles of $Na_2SO_4$ will require= $\frac{3}{3}\times 0.0076=0.0076moles$ of $H_2SO_4$

Mass of $H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g$

Thus 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

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How many moles are there in 3.612 x 1024 formula units of Na2SO4?

vitfil [10]

Answer:

<h3>Theanswer is 6 moles</h3>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{3.612 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>6 moles</h3>

Hope this helps you

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3 years ago

An orgo lab student named Bob wanted to synthesize an isoamyl acetate ester because it smells like bananas and he really likes t

mihalych1998 [28]

Answer:

The answer is "11.07 g".

Explanation:

Isoamyl alcohol is a reagent restriction

Isoamyl alcohol Moles:

$= \frac{7.5}{88.15} \\\\ =0.085$

Moles only with the shape of isoamyl acetate are equivalent to numbers.

Isoamyl acetate grams:

$= 0.085 \times 130.19\\\\ = 11.07 \ g$

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3 years ago

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Answer:

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2 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

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3 years ago

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GarryVolchara [31]

Explanation:

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2 years ago