Дано:
X - щелочной металл
m(X2O) = 3 г
m(XOH) = 4,8 г
Найти:
X -?
Решение:
X2O + H2O = 2XOH
m(X2O)/M(X2O) = m(XOH)/(2*M(XOH))
3/(2*X + 16) = 4,8/(2*(X + 16 + 1))
3/(2*X + 16) = 4,8/(2*(X + 17))
3/(2*X + 16) = 4,8/(2*X + 34)
2*X + 16 = (3*(2*X+34))/4,8
2*X + 16 = 0,625*(2*X + 34)
2*X + 16 = 1,25*X + 21,25
0,75 * X = 5,25
X = 7 - литий (Li)
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Дано:
X - щелочной металл
m(X2O) = 3 г
m(XOH) = 4,8 г
Найти:
X -?
Решение:
X2O + H2O = 2XOH
m(X2O)/M(X2O) = m(XOH)/(2*M(XOH))
3/(2*X + 16) = 4,8/(2*(X + 16 + 1))
3/(2*X + 16) = 4,8/(2*(X + 17))
3/(2*X + 16) = 4,8/(2*X + 34)
2*X + 16 = (3*(2*X+34))/4,8
2*X + 16 = 0,625*(2*X + 34)
2*X + 16 = 1,25*X + 21,25
0,75 * X = 5,25
X = 7 - литий (Li)