Ask question

Ask question

All categories

4 years ago

12

In the previous problem, which are the limiting reactants for each reaction? Based on this, calculate the maximum yield of oxyge

n, carbon dioxide, and sulfur dioxide when 280 g of black powder is burned.

1 answer:

juin [17]4 years ago

4 0

please show the previous problem

You might be interested in

What implications need to be considered when creating environmental policies?

Igoryamba

Answer:

How will it affect the environment? Is its benefit worth the cost?

Explanation:

when creating environmental policies you should always look into if it is worth it or not and how it will affect the environment

4 0

3 years ago

Read 2 more answers

Can someone please help me out?

ziro4ka [17]

Yes the answer is c because the atoms are preserved

4 0

3 years ago

Read 2 more answers

swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

5 0

4 years ago

Calculate the mass of nickel metal which will react with 25mL of 0.15mol/L (

Fittoniya [83]

Answer:

0.110625 g of Ni

Explanation:

The first step in solving this problem is to put down the accurate chemical reaction equation.

Ni(s) + 2HCl(aq) ---> NiCl2(aq) + H2(g)

Secondly, we obtain the amount of HCl that reacted from the information provided.

Volume of HCl (V)= 25 ml

Concentration of HCl (C)= 0.15 mol/L

Then, to find the number of moles of HCl (n);

n= CV

Substitution values

n= 25/1000 × 0.15

n= 3.75 ×10^-3 moles

Mass of 3.75 ×10^-3 moles of HCl = number of moles × molar mass

Molar mass of HCl= 36.5 gmol-1

Therefore;

Mass of HCl = 3.75 ×10^-3 moles × 36.5 gmol-1

Mass of HCl= 0.136875 g of HCl

Thirdly we determine the mass of Ni reacted;

If 1 mole of Ni reacted with 2 moles of HCl according to the reaction equation

Then x moles of Ni reacts with 3.75 ×10^-3 moles of HCl

x= 1 × 3.75 ×10^-3 moles/ 2

x= 1.875 × 10^-3 moles of Ni

Mass of Ni= 1.875 × 10^-3 moles of Ni × 59 gmol-1

Mass of Ni= 0.110625 g of Ni

6 0

4 years ago

You titrate 25.0 mL of HCl solution of an unknown concentration with a 0.190 M NaOH solution. Complete neutralization of the HCl

galina1969 [7]

A titration is defined as 'the process of determining the quantity of a substance A by adding measured increments of substance B, the titrant, with which it reacts until exact chemical equivalence is achieved (the equivalence point)'.

The Molarity of the HCl Solution requires the addition of 24.6 mL of titrant is 0.0836 M (to 3 significant figures)

<h3>What is Titrant?</h3>

A reagent solution of precisely known concentration that is added in titration

Given reaction HCl + NaOH ===> NaCl + $H_2O$ neutralization reaction

Note that the mole ratio is 1:1 meaning that 1 mole HCl reacts with 1 mole NaOH to produce 1 mole NaCl and 1 mole $H_2O$

Find moles of NaOH used:

18.45 ml x 1 L/1000 ml x 0.085 mol/L = 0.002091 moles NaOH used

This meant that there must be 0.002091 moles of HCl present in the 25.0 mls.

We want to find the molarity (M) which is moles HCl/liter of solution. We know the moles, and we also know the liters of solution (25.0 mls x 1 L / 1000 mls = 0.025 liters)

M = 0.002091 moles HCl / 0.025 liters = 0.08364 M = 0.0836 M (to 3 significant figures)

Learn more about Titration Here;
brainly.com/question/27107265

#SPJ1

4 0

2 years ago