N2 + 3H2 → 2NH3 How many moles of NH3 could you make with 4.2021 g of N2?

2. 2Na + Cl2 → 2NaCl

How many moles of Cl2 do you need to use in order to react with 34.485 g of Na?

3. H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4

How many grams of NaNO2 do you need in order to make 24.14714 g of Na2SO4?

4. 2Na + 2H2O → 2NaOH + H2

How many grams of NaOH could you make with 8.0465 g of Na?

5. 2Mg + O2 → 2MgO

If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

6. FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl

How many grams of NaCl could you make with 2.43306 g of FeCl3?

Step-by-step explanation:

1. N2 + 3H2 → 2NH3

How many moles of NH3 could you make with 4.2021 g of N2?

From the reaction;

1 mol of N2 produces 2 mol of NH3

1 mol of N2 = 14g of N2

14g of N2 produces 2 mol of NH3

4.2021 g of N2 produces x mol of NH3

14 = 2

4.2021 = x

x = 4.2021  * 2 / 14 = 0.6003 mol of NH3

2. 2Na + Cl2 → 2NaCl

How many moles of Cl2 do you need to use in order to react with 34.485 g of Na?

From the equation of the reaction;

2 mol of Na reacts with 1 mol of Cl2

1 mol of Na = 23g of Na

2 mol of Na - 46g of Na

46g of Na reacts with 1 mol of Cl2

4.485 g of Na reacts with x mol of Cl2

46 = 1

4.485 = x

x = 4.485 * 1 / 46 = 0.0975 mol of Cl2

3. H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4

How many grams of NaNO2 do you need in order to make 24.14714 g of Na2SO4?

From the reaction;

2 mol of NaNO2 produces 1 mol of Na2SO4

2 mol of NaNO2 = 68.995 g/mol * 2mol = 137.99g

1 mol of Na2SO4 = 142.04 g

137.99g of NaNO2 produces 142.04 g of Na2SO4

xg of NaNO2 produces 24.14714 g of Na2SO4?

137.99 = 142.04

x = 24.14714

x = 24.14714 * 137.99 / 142.04 = 23.459g

4. 2Na + 2H2O → 2NaOH + H2

How many grams of NaOH could you make with 8.0465 g of Na?

From the reaction;

2 mol of Na produces 2 mol of NaOH

2 mol of Na = 23g/mol * 2mol = 46g

2 mol of NaOH = 40g/mol * 2 = 80g

46 g of Na produces 80g of NaOH

8.0465 g of Na produces xg of NaOH

46 = 80

8.0465 = x

x = 8.0465 * 80 / 46 = 13.994g

5. 2Mg + O2 → 2MgO  

If you are burning 5.8332 g of Mg, how many grams of MgO will this make?

From the reaction;

2 mol of Mg produces 2 mol of MgO

2 mol of Mg = 24 g/mol * 2mol = 48g

2 mol of MgO = 40.3 g/mol  * 2mol = 80.6g

48g of Mg produces 80.6 g  of MgO

5.8332 g of Mg produces xg of MgO

48 = 80.6

5.8332 = x

x = 5.8332 * 80.6 / 48   = 9.8g

6. FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl

How many grams of NaCl could you make with 2.43306 g of FeCl3?

From the reaction;

1 mol of FeCl3 produces 3 mol of NaCl

1 mol of FeCl3 = 162.2 g

3 mol of NaCl = 58.44 g/mol * 3 mol = 175.32 g

162.2g of FeCl3 produces 175.32 g of NaCl

2.43306 g of FeCl3 produces xg of NaCl

162.2 = 175.32

2.43306 = x

x = 2.43306 * 175.32 / 162.2 = 2.63g

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.