Answer
$mass(g) = 0.0104$
Work Step by Step
1. Calculate [OH-]:
pH + pOH = 14
10 + pOH = 14
pOH = 4
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 4}$
$[OH^-] = 1 \times 10^{- 4}M$
$- SrO(aq) + H_2O(l) -- \gt Sr^{2+}(aq) + 2OH^-(aq)$
2. Since $SrO$ is a strong base that produces 2 OH for each molecule: $[OH^-] = 2 * [SrO]$
$ 1\times 10^{- 4} = 2 * [SrO]$
$ \frac{ 1\times 10^{- 4}}{ 2} = [SrO]$
$ 5\times 10^{- 5}M = [SrO]$
3. Calculate the number of moles:
$n(moles) = concentration(M) * volume(L)$
$n(moles) = 5\times 10^{- 5} * 2$
$n(moles) = 1\times 10^{- 4}$
4. Find the mass value in grams:
(87.62* 1 + 16* 1) = 103.62g/mol
$mass(g) = mm(g/mol) * n(moles)$
$mass(g) = 103.62 * 1\times 10^{- 4}$
$mass(g) = 0.01036$
