How many grams of chromium are needed to react with an excess of CuSo4 tp produce 27 g Cu?
5 answers
Just so the chem teachers know what your school subject is...
1. Write the equation and balance it.
2. Convert 27 g Cu to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Cu to moles Cr.
4. Now convert moles Cr to grams. g = moles x molar mass.
2. Convert 27 g Cu to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Cu to moles Cr.
4. Now convert moles Cr to grams. g = moles x molar mass.
14.7
help w this pls
Solution:
The chemical equation for this reaction is: 3CuSO4 + 2Cr = Cr2(SO4)3 + 3Cu
The amount of Cu produced in reaction is n(mol) = m(g) / MW(g/mol)
n(Cu) = 27 / 63.5 = 0.425 mol
According to the chemical equation n(Cr) = 2⁄3 ∙ n(Cu)
n(Cr) = 2⁄3 ∙ 0.425 = 0.283 mol
The mass of Cr is
m(g) = n(mol) ∙ MW(g/mol)
The molar weight of Cr is equal to its atomic weight in the periodic table of elements (MW(Cr) = 52 g/mol)
m(Cr) = 0.283 ∙ 52 = 14.7 g Answer: m(Cr) = 14.7 g
The chemical equation for this reaction is: 3CuSO4 + 2Cr = Cr2(SO4)3 + 3Cu
The amount of Cu produced in reaction is n(mol) = m(g) / MW(g/mol)
n(Cu) = 27 / 63.5 = 0.425 mol
According to the chemical equation n(Cr) = 2⁄3 ∙ n(Cu)
n(Cr) = 2⁄3 ∙ 0.425 = 0.283 mol
The mass of Cr is
m(g) = n(mol) ∙ MW(g/mol)
The molar weight of Cr is equal to its atomic weight in the periodic table of elements (MW(Cr) = 52 g/mol)
m(Cr) = 0.283 ∙ 52 = 14.7 g Answer: m(Cr) = 14.7 g
1 answer