Answer: -
37.5 g of NaCl are in 250 g of a 15.0% (by weight) solution.
Explanation: -
Total mass = 250 g
Percentage of NaCl = 15.0 %
=
Thus =
Number of grams of NaCl present =
= 37.5 g
Answer:
single replacement
Explanation:
Step 1: Data given
single replacement = A reaction in which one element replaces a similar element in a compound. For example, a metal replaces an other metal.
The general form of a single-replacement (also called single-displacement) reaction is:
A+BC→AC+B
Decomposition = a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:
AB→A+B
Synthesis = A reaction that occurs when one or more compounds combines to form a complex compound:
A + B → AB
Double replacement: a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds.
The general form of a double-replacement reaction is:
AB+CD→AD+BC
Combustion reaction = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
The reaction Zn + 2HCl → ZnCl2 + H2
⇒ Does not involve O2 = NOT a combustion reaction
⇒ The compounds do not form a complex compound = NOT a synthesis
⇒ A compound does not break down into smaller substances = NOT a decomposition
⇒ There is a replacement between Zn and H. This is a <u>single replacement</u>, not a double replacement reaction.
Answer:
71.5g
Explanation:
The reaction equation is given as:
C + O₂ → CO₂
Mass of C = 42g
Mass of O₂ = 52g
Unknown:
Mass of CO₂ produced = ?
Solution
Now to solve this problem, we have to find limiting reactant which is the one given in short supply in this reaction.
The extent of the reaction is controlled by this reactant.
Find the number of moles of the given species;
Number of moles =
Number of moles of C = = 3.5mol
Number of moles of O₂ = = 1.63mol
Now;
From the balanced reaction equation;
1 mole of C reacted with 1 mole of O₂
We see that C is in excess and O₂ is the limiting reactant.
1 mole of O₂ will produce 1 mole of CO₂
So; 1.63mole of O₂ will produce 1.63 mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 44g/mol
Mass of CO₂ = 1.63 x 44 = 71.5g