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Papessa [141]
4 years ago
14

Identify the limiting reactant when 6.45 g of O2 gas reacts with 1.13 g of H2 gas to produce liquid water.

Chemistry
1 answer:
sergeinik [125]4 years ago
3 0

Answer:

Oxygen is the limiting reactant

Explanation:

2H2 +    O2 ⇒     2H2O

2mol      1mol        2mol

2moles of hydrogen will require 1mol of oxygen to produce 2mol of water

in the question, 6.45g of O2 reacts with 1.13g of H2.

molar mass of O2 = 16*2 =32g/mol

molar mass of H2 = 1*2=2g/mol

number of moles of O2 = mass/molar mass

                                      = 6.45/32

                                     =0.20156moles

number of moles of H2 = 1.13/2

                                   =0.565moles

but 0.20156moles of O2 will require 0.40312moles of hydrogen

2H2          +        O2        ⇒       2H2O

2mol                  1mol                 2mol

0.40312mole    0.20156mol

0.40312moles is less than 0.565moles of H2 present, which means hydrogen is in excess hence implying that oxygen is the limiting reactant

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The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
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Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

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Now, it is given that:

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Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

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