Answer:
88%.
Explanation:
The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
P = 14.9258934 atm
Answer: 40.3 L
Explanation:
To calculate the moles :
According to stoichiometry :
1 moles of
produces = 3 moles of
Thus 0.600 moles of
will produce=
of
Volume of
Thus 40.3 L of CO is produced.
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag