Hi, please show the step process in solving the following equations with the ans

Question

Hi, please show the step process in solving the following equations with the answers. Thanks for your continued support.

1.) Find the mass of glucose (C6H12O6) required to prodice 1.82L of carbon dioxide gas at STP from the reaction. C6H12O6 ->2C2H6O + 2CO2.

2.) How many liters of oxygen react with sulfur to produce 425 g of sulfur dioxide?

3.) Find the mass of sodium required to produce 5.68 L of hydrogen gas at STP from the reaction: 2 Na + O2 -> 2NaOH + H2.

4.) How many liters of hydrogen are produced if 225 g of iron reacts with hydrochloric acid in a single replacement reaction.

5.) How many liters of oxygen are necessary for the combusion of 134 g of magnesium into magnesium oxide.

6.) What volume of oxygen gas is produced by the decomposition of 100.0 grams of sodium nitrate? NaNO3 -> NaNO2 + O2

7.) What volume of hydrogen is produced when 75.0 grams of water is decomposed by electrolysis?

Explanation / Answer

a. moles CO2 = 1.82 L/ 22.4 L/mol= 0.0812

the ratio between C6H12O6 and CO2 is 1 : 2

moles C6H12O6 required = 0.0812 x 1/2=0.0406

mass C6H12O6 = 0.0406 mol x 180 g/mol=7.31 g

2. moles S = 425 g / 32.066 g/mol=13.2

the ratio between S and O2 is 1 : 1

moles O2 required = 13.2

Volume of O2 at STP = 13.2 mol x 22.4 L/mol=295.7 L

3. At STP one mole of any gas occupies 22.4 L

moles H2 = 5.68 /22.4 =0.253

the ratio between Na and H2 is 2 : 1

moles Na = 2 x 0.253 =0.506

mass Na = 0.506 mol x 22.9898 g/mol=11.6 g

4. Fe+2HCL = FeCl2 +H2

Iron moles = 225/55.8 = 4.0322

HenceH2 produced = 4.0322 moles.

At STP Volume of H2 =22.4*4.0322=90.322 L

5.

2 Mg + O2 ? 2 MgO

(134 g Mg) / (24.30506 g Mg/mol) x (1/2) x (22.414 L/mol) = 61.8 L O2 at STP

6.

Since the equation is balanced we can use the coefficients as ratios. the ratios for question 1 are. 2 : 2 : 1

Step 1:
We have to convert the mass of sodium nitrate to moles, by using the formula
n=m/M
n=100 g/ 85 (molar mass of sodium nitrate)
n=1.176 mol

Step 2:
We have to find the number of moles in O2, using the ratios and the number of moles of sodium nitrate.
number of moles of sodium nitrate multiplied by the ratios (coefficient of unknown/coefficient of known)

n(O2)=n(NaNO3) x 1/2
n(O2)=1.176 x 1/2
n(O2)=0.588 mol

Step 3:
We now have to convert the number of moles of O2 to a volume in liters.
Since we assumed that this reaction is at STP, we are going to use the formula

V= n x 22.41

So

V(O2)= 0.588 x 22.41
V(O2)= 13.18 liters

7.

Also assuming reaction is at STP

The equation is balanced and the coefficients can be used as ratios

ratios are 2 : 2 : 1

Step 1:

Find the number of moles of H2O by using the formula n=m/M

n=75g/18.02
n=4.16 mol

Step 2:
Find the number of moles of oxygen by using the ratios and the number of moles of water.
Number of moles of water multiplied by (coefficient of unknown/coefficient of known)

n(O2)=n(H2O) x 1/2
n(O2)=4.16 mol x 1/2
n(O2)=2.08 mol

Step 3:
Convert the n(O2) to the volume using the formula: V= n x 22.41
V(O2)= 2.08 x 22.41
V(O2)= 46.6 liters

V(H2) 2*46.6 = 93.2 L

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