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What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0

3 years ago

Question 1 How many grams are equal to 3.4 moles of carbon dioxide?

ddd [48]

Answer:

149.6 grams

Explanation:

Mass in gram = molar mass * number of moles

Massof CO2 in gram = 44*3.4=149.6 grams

4 0

2 years ago

Which metal will react spontaneously with Cu2+ (aq) at 25°C?

Gwar [14]

Answer:

Mg

Explanation:

The standard reduction potentials are

E°/V

Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42

Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85

Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80

Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34

Mg2+(aq) + 2e- ⟶ Mg(s); -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

5 0

2 years ago

Read 2 more answers

A student experimentally obtained the density of osmium the densest element as 22.57g/cm3 the density of osmium is reported to b

Airida [17]

Answer:

The answer to your question is: % error = 0.4

Explanation:

Data

real value = 22.48%

estimated value = 22.57 %

Formula

% error = |real value - estimated value|/real value x 100

%error = |22.48 - 22.57|/22.48 x 100

% error = |-0.09|/22.48 x 100

%error = 0.09/22.48 x 100

% error = 0.004 x 100

% error = 0.4

3 0

3 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago