Question

2C4H10 + 13O2  →  10H2O + 8CO2

What is the molar mass of O2 ? 

Question 1 options:

416.00 g/mol

208.00 g/mol

16.00 mol/g

32.00 mol/

Given the balanced equation:

2C4H10 + 13O2  →  10H2O + 8CO2

What is the molar mass of C4H10 ? 

Question 2 options:

116.28 g/mol

58.14 g/mol

116.28 mol/

Question 3 (2 points)

 

Saved

Given the balanced equation:

2C4H10 + 13O2  →  10H2O + 8CO2

If you are given 150 grams of C4H10 ,How many moles would that be? 

Question 3 options:

5.16 mol

2.58 mol

0.19 mol

0.39 mo

Question 4 (3 points)

 

Given the balanced equation:

2C4H10 + 13O2  →  10H2O + 8CO2

If you are given 150 grams of C4H10 ,How many moles of O2 would you need to completely react with the C4H10 ? 

Question 4 options:

5.16 mol

8.39 mol

16.77 mol

0.06 mo

Question 5 (4 points)

 

Given the balanced equation:

2C4H10 + 13O2  →  10H2O + 8CO2

If you are given 150 grams of C4H10 ,How many grams of O2 would you need to completely react with the C4H10 ? 

Question 5 options:

150

975 grams

4800

537 gram

Answer 1

Ans:- (1)

Molar mass of Oxygen is 16 g/mol

Mass of $O_2$ = 16$\times$ 2 $\Rightarrow$ 32g/mol

So the molar mass of $O_2$ according to the balanced equation is =13$\times$ 32 $\Rightarrow$416.00g/mol

Ans:- (2)

Molecular mass of $C_4H_{10}$ = 58.14g/mol

So the molar mass of $C_4H_{10}$ according to balanced equation is = 58.14$\times$ 2$\Rightarrow$ 116.28 g/mol

Ans:- (3)

Moles of $C_4H_{10}$ = $\dfrac{150}{58.14}$ $\Rightarrow$ 2.58 mol

So, the moles of $C_4H_{10}$ according to stoichiometry is 2.58$\times$ 2 $\Rightarrow$ 5.16 mol

Ans:-(4)

According to Stoichiometry

Since, 2 moles of $C_4H_{10}$ = 13 moles of $O_2$ reacted

moles of $O_2$ reacted completely to produce product= $\dfrac{13\times 2.58}{2}=16.77mol$

Ans:-(5)

Grams of $O_2$ = moles of $O_2 \times$ molar mass of $O_2$

$\Rightarrow$${16.77}\times32$

$\Rightarrow$ 537 gm