Answer: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Explanation:
Molar mass ( CuSO₄) = 159.609 g/mol
159.609 g ----------------- 6.02 x 10²³ molecules
? g ------------------ 3.36 x 10²³ molecules
mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³
mass = 5.36 x 10²⁴ / 6.02 x 10²³
mass = 8.90 g
hope this helps!
Answer:
33.33% = 33%
Explanation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
1 mole of MCO3 will produce → 1 mole of CO2
We need to get the number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
Moles = 0.22 g / 44 g/mol = 0.005 mole
Moles of Mg = moles of CO2 = 0.005 mole
Mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
=33.33 %
Answer:
Total mass = 246 g
Explanation:
Given data:
Mass of vinegar = 59 g
Mass of oil = 177 g
Mass of brown sugar = 10 g
Total mass = ?
Solution:
Total mass = masses of [sugar + oil + vinegar]
Total mass = 59 g + 177 g + 10 g
Total mass = 246 g